KINEMATICS REVIEW LESSON EXAMPLE 1: Consider the velocity vs. time graph below. The motion of five different people (A, B, C, D, and E) are shown on the graph. Study the graph and answer the following questions. Some questions may have more than or less than one answer. Question: 1. Which person(s) changes direction during the time of motion? 2. Which person(s) is/are not moving? 3. Which person has the greatest average speed? 4. Which person(s) has/have a constant, positive acceleration value? 5. Which person has the greatest magnitude of acceleration? 1. C 2. none 3. E 4. D reason: lets say initial vel = -100, and final velocity = -50 -> a = v f – v i /t = -50 – (-100) = +50/t 5. C (largest change in velocity) EXAMPLE 2: An engineer is designing the runway for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3.0 m/s 2 . The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway? v f 2 = v i 2 + 2*a*d
Transcript
KINEMATICS REVIEW LESSON
EXAMPLE 1: Consider the velocity vs. time graph below. The motion of five different people (A, B, C, D, and E) are shown on the graph. Study the graph and answer the following questions. Some questions may have more than or less than one answer.
Question:
1. Which person(s) changes direction during the time of motion?
2. Which person(s) is/are not moving? 3. Which person has the greatest average speed? 4. Which person(s) has/have a constant, positive
acceleration value?
5. Which person has the greatest magnitude of acceleration?
1. C
2. none
3. E
4. D reason: lets say initial vel = -100, and final velocity = -50 -> a = vf – vi/t = -50 – (-100) = +50/t
5. C (largest change in velocity)
EXAMPLE 2:
An engineer is designing the runway for an airport. Of the planes which will use the airport, the lowest acceleration rate is likely to be 3.0 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?
vf2 = vi
2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m2/s2)/(6 m/s2) = d
d = 704 = 7.0 x 102 m
EXAMPLE 3: A train leaves the station traveling North at 55 km/h. Leaving at the same instant, a train leaves the station traveling South at 65 km/h. The distance between the two train stations is 150 km. When and where do they meet?
Recall, that when two velocities are opposite directions they have to have opposite signs, therefore we will call the velocity going South, negative (-65 km/h).
Draw a position-time graph, using the slopes of the lines, you can roughly determine when the two trains will pass.
dnorth = dsouth at catch up points
Since the train leaving the South is designated as having a negative velocity, it starts at 150 km point and the train leaving the North starts at 0 (as the slope of the velocity is positive)
dsouth = vsoutht + do
dnorth = vnortht
vsoutht + do = vnortht
-65t + 150 = 55t
t = 1.25 = 1.3 hours
Substitute t into one of the equations to find the distance when they pass:
dsouth = vsoutht + do = -65(1.25) + 150 = 68.75 = 69 km from the North
Why is it from the North? (look to graph for answer, what does d represent)
EXAMPLE 5: A ball is thrown vertically upward from a window that is 3.6 m above the ground. The ball's initial speed is 2.8 m/s and the acceleration due to gravity is 9.8 m/s2. (a) What is the ball's speed when it hits the ground? (b) How long after the first ball is thrown should a second ball be simply dropped from the same window so that both balls hit the ground at the same time?
NOTE: Let down be the "negative" direction and up be the "positive" direction.
a. vf = (vi2 + 2ad)1/2
vf = [2.82 + 2(-9.8)(3.6)]1/2
vf = 8.9 m/s
b. The ball's speed is 8.9 m/s.
(a) first ball:
t = (vf – vi)/a
= -8.9 –(2.8)/-9.8 = 1.2 s
second ball:
t = [2d/a]1/2
=[2(0-3.6)/-9.8]1/2
= 0.85 s
Time to wait: t = 1.19 s – 0.85 s = 0.3 s
EXAMPLE 6:
Arthur's catapult launches a 90 kg cannonball over Henry the Red's castle wall. The launch speed of the cannonball is 25 m/s, at an angle of 35 degrees from the horizontal. Ignore air resistance, and assume that the ground is level.
a. What are the initial x & y components of the projectile's velocity?
b. What are the x & y accelerations for the projectile?
c. What is the maximum height of the projectile? Explain the theoretical assumption and the calculations you make to find this.
d. What is the total time of flight for this projectile? Explain the theoretical assumption and the calculation you make to find this.
e. What is the total range traveled by this projectile? Explain the theoretical assumption and
the calculation you make to find this.
a. What are the initial x & y components of the projectile's velocity?
We can use trigonometry to solve for the initial x & y components of the projectiles velocity because we are given the angle 35 degrees above the horizontal and the hypotenuse of the triangle. To solve for the initial y component of the velocity, we use the sine function as follows:
Sin 35° = y/(25m/s)
y = (25m/s)(sin35° ) = 14.3m/s
To solve for the initial x component of the velocity, we use the cosine function as follows:
Cos 35° = x/(25m/s)
x = (25m/s)(cos35° ) = 20.4m/s
b. What are the x & y accelerations for the projectile?
The acceleration in the y direction is equal to the gravitational field strength of the earth, which is -9.8m/s2.
The acceleration in the x direction is equal to 0m/s2 because we have assumed no air resistance.
c. What is the maximum height of the projectile? Explain the theoretical assumption and the calculations you make to find this.
In order to solve for the maximum height of the projectile (ymax), we must first solve for the time of flight of the projectile. Proceed to question d. first and then solve for ymax.
It turns out that the travel time of the projectile is 2.92 seconds. However, we are interested in the time when the cannonball is at the highest point in it's trajectory. This
occurs at t1/2, or 1.46 seconds. We must employ the kinematics equation: x = vot + 1/2at2
y1/2 = (voy)(t1/2) + 1/2(ay)(t2)
y1/2 = (14.3m/s)(1.46s) +1/2(-9.8m/s2)(1.46)2
y1/2 = 10.4 meters
d. What is the total time of flight for this projectile? Explain the theoretical assumption and the calculation you make to find this.
At the instant that the cannonball is at the top of it's trajectory (when it has stopped rising, but before it begins it's desent), it's velocity in the y direction becomes zero. Because the trajectory of a parabola is symmetrical with respect the the y-axis, we know that this moment occurs halfway through the cannonball's flight (t1/2).We can use this very important piece of information to solve for the time of flight of the cannonball. To do this we must employ the kinematics equation: v = vo + at
v1/2y = voy + (ay)(t1/2)
(v1/2y - voy)/(ay) = t1/2
(0 m/s - 14.3 m/s)/(-9.8m/s2) = t1/2
t1/2 = 1.46
Here, we have solved for the time it takes the cannonball to complete half of it's flight. To solve for the total flight time, we must multiply this t1/2 by 2.
total flight time = 2(t1/2) = 2.92 seconds
e. What is the total range traveled by this projectile? Explain the theoretical assumption and the calculation you make to find this.
To solve for the range of the cannonball, we can employ the definition: distance = rate x time. To make this work out, we have to make the assumption that there is no air resistance. This ensures that the velocity in the x direction stays constant.
We already solved for the initial velocity of the cannonball in the x direction and the total flight time. The sloution is as follows:
distance = rate x time
d = (20.4 m/s)(2.92 s) = 59.6 meters
Kinematics Review vfy
2 = viy2 + 2ay Δd, where Δd = df - di
vfy = viy + ay*t
Δd = vi*t + 0.5*a*t2, where Δd = df - di
ONE DIMENSIONAL KINEMATICS
1. Which of the following statements about vectors and scalars are TRUE? List all that apply.
a. A vector is a large quantity and a scalar is a small quantity. b. A scalar quantity has a magnitude and a vector quantity does not. c. A vector quantity is described with a direction and a scalar is not. d. Scalar quantities are path dependent quantities and vector quantities are not. e. A scalar quantity depends only upon the initial and final values of the quantity;
this is not the case for vector quantities. f. The quantity 20 m/s, north is a speed and as such is a scalar quantity. g. The quantity 9.8 m/s/s is an acceleration value and as such is a vector quantity.
2. Which of the following statements about distance and/or displacement are TRUE? List all that apply.
a. Distance is a vector quantity and displacement is a scalar quantity. b. A person makes a round-trip journey, finishing where she started. The
displacement for the trip is 0 and the distance is some nonzero value. c. A person starts at position A and finishes at position B. The distance for the trip is
the length of the segment measured from A to B. d. If a person walks in a straight line and never changes direction, then the distance
and the displacement will have exactly the same magnitude. e. The phrase "20 mi, northwest" likely describes the distance for a motion. f. The phrase "20 m, west" likely describes the displacement for a motion. g. The diagram below depicts the path of a person walking to and fro from position
A to B to C to D. The distance for this motion is 100 yds. h. For the same diagram below, the displacement is 50 yds.
3. Which of the following statements about velocity and/or speed are TRUE? List all that apply.
a. Velocity is a vector quantity and speed is a scalar quantity. b. Both speed and velocity refer to how fast an object is moving. c. Person X moves from location A to location B in 5 seconds. Person Y moves
between the same two locations in 10 seconds. Person Y is moving with twice the speed as person X.
d. The velocity of an object refers to the rate at which the object's position changes. e. For any given motion, it is possible that an object could move very fast yet have
an abnormally small velocity. f. The phrase "30 mi/hr, west" likely refers to a scalar quantity. g. The average velocity of an object on a round-trip journey would be 0. h. The direction of the velocity vector is dependent upon two factors: the direction
the object is moving and whether the object is speeding up or slowing down. i. The diagram below depicts the path of a person walking to and from position A to
B to C to D. The entire motion takes 8 minutes. The average speed for this motion is approximately 11.3 yds/min.
j. For the same diagram below, the average velocity for this motion is 0 yds/min.
4. Which of the following statements about acceleration are TRUE? List all that apply.
a. Acceleration is a vector quantity. b. Accelerating objects MUST be changing their speed. c. Accelerating objects MUST be changing their velocity. d. Acceleration units include the following; m/s2, mi/hr/sec, cm/s2, km/hr/m.
e. The direction of the acceleration vector is dependent upon two factors: the direction the object is moving and whether the object is speeding up or slowing down.
f. An object which is slowing down has an acceleration. g. An object which is moving at constant speed in a circle has an acceleration. h. Acceleration is the rate at which the velocity changes. i. An object that is accelerating is moving fast. j. An object that is accelerating will eventually (if given enough time) be moving
fast. k. An object that is moving rightward has a rightward acceleration. l. An object that is moving rightward and speeding up has a rightward acceleration. m. An object that is moving upwards and slowing down has an upwards acceleration.
5. Which of the following statements about position-time graphs are TRUE? List all that apply.
a. Position-time graphs cannot be used to represent the motion of objects with accelerated motion.
b. The slope on a position-time graph is representative of the acceleration of the object.
c. A straight, diagonal line on a position-time graph is representative of an object with a constant velocity.
d. If an object is at rest, then the position-time graph will be a horizontal line located on the time-axis.
e. Accelerated objects are represented on position-time graphs by curved lines. f. An object with a positive velocity will be represented on a position-time graph by
a line with a positive slope. g. An object with a negative velocity will be represented on a position-time graph by
a line with a negative slope. h. An object with a positive acceleration will be represented on a position-time
graph by a line which curves upwards. i. An object with a negative acceleration will be represented on a position-time
graph by a line which curves downwards.
6. Which of the following statements about velocity-time graphs are TRUE? List all that apply.
a. The slope on a velocity-time graph is representative of the acceleration of the object.
b. The area on a velocity -time graph is representative of the change in position of the object.
c. An accelerated object's motion will be represented by a curved line on a velocity-time graph.
d. Objects with positive acceleration will be represented by upwardly-curved lines on a velocity-time graph.
e. If an object is at rest, then the velocity-time graph will be a line with zero slope.
f. A line with zero slope on a velocity-time graph will be representative of an object which is at rest.
g. A line with a negative slope on a velocity-time graph is representative of an object with negative velocity.
h. If an object changes its direction, then the line on the velocity-time graph will have a changing slope.
i. An object which is slowing down is represented by a line on a velocity-time graph which is moving in the downward direction.
7. Which of the following statements about free fall and the acceleration of gravity are TRUE? List all that apply.
a. An object that is free-falling is acted upon by the force of gravity alone. b. A falling skydiver which has reached terminal velocity is considered to be in a
state of free fall. c. A ball is thrown upwards and is rising towards its peak. As it rises upwards, it is
NOT considered to be in a state of free fall. d. An object in free fall experiences an acceleration which is independent of the
mass of the object. e. A ball is thrown upwards, rises to its peak and eventually falls back to the original
height. As the ball rises, its acceleration is upwards; as it falls, its acceleration is downwards.
f. A ball is thrown upwards, rises to its peak and eventually falls back to the original height. The speed at which it is launched equals the speed at which it lands. (Assume negligible air resistance.)
g. A very massive object will free fall at the same rate of acceleration as a less massive object.
h. The value of g on Earth is approximately 9.8 m/s2. i. The symbol g stands for the force of gravity.
8. If an object has an acceleration of 0 m/s2, then one can be sure that the object is not ____.
a. moving b. changing position c. changing velocity
9. If car A passes car B, then car A must be ____.
a. accelerating. b. accelerating at a greater rate than car B. c. moving faster than car B and accelerating more than car B. d. moving faster than car B, but not necessarily accelerating.
10. Which one of the following is NOT consistent with a car which is accelerating?
a. A car is moving with an increasing speed.
b. A car is moving with a decreasing speed. c. A car is moving with a high speed. d. A car is changing direction.
11. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. This is evidence that
a. he is accelerating b. he is covering a greater distance in each consecutive second. c. he is moving with a constant speed (on average).
12. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. What is the player's acceleration?
13. Olympic gold medalist Michael Johnson runs one time around the track - 400 meters - in 38 seconds. What is his displacement? ___________ What is his average velocity? ___________
14. If an object is moving eastward and slowing down, then the direction of its velocity vector is ____.
a. eastward b. westward c. neither d. not enough info to tell
15. If an object is moving eastward and slowing down, then the direction of its acceleration vector is ____.
a. eastward b. westward c. neither d. not enough info to tell
16. Which one of the following quantities is NOT a vector?
a. 10 mi/hr, east b. 10 mi/hr/sec, west c. 35 m/s, north d. 20 m/s
17. Which one of the following quantities is NOT a speed?
a. 10 mi/hr b. 10 mi/hr/sec c. 35 m/s d. 20 m/s
18. Which one of the following statements is NOT true of a free-falling object? An object in a state of free fall ____.
a. falls with a constant speed of -10 m/s. b. falls with a acceleration of -10 m/s/s. c. falls under the sole influence of gravity. d. falls with downward acceleration which has a constant magnitude.
19. The average speed of an object which moves 10 kilometers (km) in 30 minutes is ____.
a. 10 km/hr b. 20 km/hr c. 30 km/hr d. more than 30 km/hr
20. What is the acceleration of a car that maintains a constant velocity of 55 mi/hr for 10 seconds?
a. 0 b. 5.5 mi /hr/s c. 5.5 mi /s/s d. 550 mi/hr/s
21. As an object freely falls, its ____.
a. speed increases b. acceleration increases
c. both of these d. none of these
22. A speedometer is placed upon a free-falling object in order to measure its instantaneous speed during the course of its fall. Its speed reading (neglecting air resistance) would increase each second by about ____.
a. 5 m/s. b. 10 m/s. c. 15 m/s. d. a variable amount. e. depends on its initial speed.
23. Ten seconds after being dropped from rest, a free-falling object will be moving with a speed of about ____.
a. 10 m/s. b. 50 m/s. c. 100 m/s. d. more than 100 m/s.
24. A baseball pitcher delivers a fast ball. During the throw, the speed of the ball increases from 0 to 30 m/s over a time of 0.10 seconds. The average acceleration of the baseball is ____ m/s2.
a. 3 b. 30 c. 300 d. 3000 e. none of these
25. On takeoff, a rocket accelerates from rest at a rate of 50 m/s2 for 1 minute. The rocket's speed after this minute of steady acceleration will be ____. m/s
a. 50 b. 500 c. 3000 d. 3600 e. none of these
26. When a rock is dropped, it will accelerate downward at a rate of 9.8 m/s2. If the same rock is thrown downward (instead of being dropped from rest), it acceleration will be ____. (Ignore air resistance effects.)
a. less than 9.8 m/s2 b. 9.8 m/s2 c. more than 9.8 m/s2
27. Consider drops of water that leak from a dripping faucet at a constant rate. As the drops fall they ____.
a. get closer together. b. get farther apart.
c. remain at a relatively fixed distance from one another.
28. Renatta Oyle is again found driving her '86 Yugo down Lake Avenue, leaving the following trail of oil drops on the pavement.
If her car is moving from right to left, then ...
a. her velocity has a rightward direction and her acceleration has a rightward direction.
b. her velocity has a rightward direction and her acceleration has a leftward direction.
c. her velocity has a leftward direction and her acceleration has a rightward direction.
d. her velocity has a leftward direction and her acceleration has a leftward direction.
29. On the diagrams below, construct a dot diagram representing the motion of an object with a ... .
a. constant rightward velocity b. rightward velocity and a rightward acceleration c. rightward velocity and a leftward acceleration d. rightward velocity, first slow and constant, and then accelerating to a high speed e. rightward velocity, first decelerating from a high speed to a rest position, then
maintaining the rest position, and finally accelerating at a lower rate than the initial deceleration.
30. On a dot diagram, how does the motion of an object moving to the right and slowing down differ from an object moving to the left and speeding up? Explain.
31. On the position-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.)
a. at rest. b. moving in the positive direction with constant speed c. moving in the negative direction and speeding up d. moving in the positive direction and slowing down e. moving in the positive direction at a constant speed (slow) and then later fast at
constant speed f. moving with a negative velocity and a negative acceleration g. moving with a negative velocity and a positive acceleration
32. On the velocity-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.)
a. at rest b. moving in the positive direction at constant speed c. moving in the negative direction from slow to fast d. moving in the negative direction from fast to slow e. moving with a positive velocity and a positive acceleration f. moving with a positive velocity and a negative acceleration g. moving with a positive velocity at constant speed and then decelerating to a rest
position h. moving in the positive direction while slowing down, changing directions and
moving in the negative directions while speeding up
The velocity-time graph below depicts the motion of an automobile as it moves through Glenview during rush hour traffic. Use the graph to answer questions #35 - #39.
33. Determine the displacement of the automobile during the following intervals of time. PSYW
t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s
34. Determine the velocity of the automobile at the following instant(s) in time.
t = 3 s t = 8 s t = 17 s
35. Determine the acceleration of the automobile during the following intervals of time.
t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s
36. Using complete sentences and the language of physics, describe the motion of the automobile during the entire 20 seconds. Explicitly describe any changes in speed or direction which might occur; identify intervals of time for which the automobile is at rest, the automobile is moving with constant speed, or the automobile is accelerating.
37. Supposing the automobile has an oil leak, demonstrate your understanding of its motion by drawing an oil drop diagram for the 20 seconds of motion. Divide the diagram into three distinct time intervals (0-5 seconds, 5-15 seconds, 15-20 seconds).
38. For the plots below, determine the velocity of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. at 13 seconds
39. For the plots below, determine the acceleration of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. at 13 seconds
40. For the plots below, determine the displacement of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. from 0 -15 seconds
41. Determine the acceleration (in m/s2) of an object which ... .
a. moves in a straight line with a constant speed of 20 m/s for 12 seconds
b. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds c. changes its velocity from 0 mi/hr to 60.0 mi/hr in 4.20 seconds d. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m
42. Determine the magnitude of the displacement (in meters) of an object which ... .
a. moves from Hither to Yon (with an average speed of 28 m/s) and then back to Hither (with an average speed of 28 m/s) if both the forward and the return trip take 46 minutes each.
b. moves at a constant speed of 8.3 m/s in a straight line for 15 seconds. c. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s d. accelerates from rest at a rate of 3.67 m/s2 for 12.1 seconds e. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s2 for 17 seconds
43. The hare is sleeping at a location that is 1200 m from the finish line. The tortoise passes him at a steady speed of 5.0 cm/s. If the hare finally wakes up 6.5 hours later, then what minimum acceleration (assumed constant) must he have in order to pass the tortoise before the finish line.
44. A Gold Car moving at 12 m/s/s passes a Green Car while the Green Car is at rest at a stoplight. The Green Car immediately accelerates at a rate of +1.8 m/s for 11 seconds. After how much time (relative to the initial starting time) must the Green Car drive before catching up with the Gold Car.
45. Ima Rilla Saari is cruising at 28 m/s down Lake Avenue and through the forest preserve. She notices a deer jump into the road at a location 62 m in front of her. Ima first reacts to the event, then slams on her brakes and decelerates at -8.1 m/s2, and ultimately stops a picometer in front of the frozen deer. What is Ima's reaction time? (i.e., how long did it take Ima to react to the event prior to decelerating?)
46. A two-stage rocket accelerates from rest at +3.57 m/s2 for 6.82 seconds. It then accelerates at +2.98 m/s2 for another 5.9 seconds. After the second stage, it enters into a state of free fall. Determine:
a. the maximum speed b. the maximum altitude c. the height of the rocket after 20.0 seconds d. the total time the rocket is in the air (assuming it is launched from the ground)
47. In a 200.0-m relay race (each leg of the race is 50 m long), one swimmer has a 0.45 second lead and is swimming at a constant speed of 3.9 m/s towards the opposite end of the pool. What minimum speed must the second swimmer have in order to catch up with the first swimmer by the end of the pool?
48. A drag racer accelerates from rest at an average rate of +13.2 m/s2 for a distance of 100.0 m. The driver coasts for 0.50 seconds and then uses the brakes and parachute to
decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the end of the track?
TWO-DIMENSIONAL KINEMATICS
49. Which of the following statements are true of scalars and vectors? List all that are TRUE.
a. A vector quantity always has a direction associated with it. b. A scalar quantity can bave a direction associated with it. c. Vectors can be added together; scalar quantites cannot. d. Vectors can be represented by an arrow on a scaled diagram; the length of the
arrow represents the vector's magnitude and the direction it points represents the vector's direction.
50. Which of the following quantities are vectors? Include all that apply.
a. distance traveled b. displacement c. average speed d. average velocity e. instantaneous velocity f. acceleration
51. Numerical values and directions are stated for a variety of quantities. Which of these statements represent a vector description? Include all that apply.
a. 20 meters, west b. 9.8 m/s/s c. 35 mi/hr, south d. 16 years old e. 60 minutes f. 3.5 m/s/s, south g. -3.5 m/s/s h. +20 degrees C
52. Which of the following statements are true of the time of flight for a projectile? List all that apply.
a. The time that a projectile is in the air is dependent upon the horizontal component of the initial velocity.
b. The time that a projectile is in the air is dependent upon the vertical component of the initial velocity.
c. For a projectile which lands at the same height that it is projected from, the time to rise to the peak is equal to the time to fall from its peak to the original height.
d. For the same upward launch angles, projectiles will stay in the air longer if the initial velocity is increased.
e. Assume that a kicked ball in football is a projectile. If the ball takes 3 seconds to rise to the peak of its trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground.
53. An object is undergoing free fall motion. As it falls, the object's ____.
a. speed increases b. acceleration increases
c. both of these d. none of these
54. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is _______.
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
55. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its acceleration is _______. (Neglect the effects of air resistance.)
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
56. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, the net force acting upon it is _______. (Neglect the effects of air resistance.)
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
57. At what point in its path is the horizontal component of the velocity (vx) of a projectile the smallest?
a. The instant it is thrown. b. Halfway to the top. c. At the top.
d. As it nears the top. e. It is the same throughout the path.
58. At what point in its path is the vertical component of the velocity (vy) of a projectile the smallest?
a. The instant it is thrown. b. Halfway to the top. c. At the top.
d. As it nears the top. e. It is the same throughout the path.
59. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity ___.
a. decreases b. remains constant c. increases
60. A bullet is fired horizontally and hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it would have hit the ground in ____. (Assume no air resistance)
a. less than 0.5 s. b. more than 0.5 s. c. 0.5 s.
61. A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what other projection angle would produce the same down-range distance?
a. 30 degrees. b. 45 degrees. c. 50 degrees. d. 75 degrees e. 90 degrees.
62. Two projectiles are fired at equal speeds but different angles. One is fired at angle of 30 degrees and the other at 60 degrees. The projectile to hit the ground first will be the one fired at (neglect air resistance) ____.
a. 30 degrees b. 60 degrees c. both hit at the same time
63. Consider the trajectory diagram shown below for a non-horizontally launched projectile. On the diagram, draw vector arrows representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first position.)
64. Consider the diagram below. The blue path represents the trajectory of a projectile dropped from rest from the top of the path. (Each ball location represents the location
after a 1-second interval of time.) The red path represents the trajectory of the same ball thrown horizontally in the absence of gravity. Construct a third path accurately showing the trajectory of a projectile launched with the same horizontal speed as the red ball. Show the location during the first four seconds of motion. Finally, the (x, y) coordinate position of location 1 is (0 m, -5 m) and the (x, y) coordinate position of location 2 is (15 m, 0 m). Determine the (x, y) coordinate positions of the four locations in your trajectory. (Assume g ~10 m/s/s.)
65. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6-meter high building. Determine the horizontal displacement of the projectile.
66. A projectile is launched with an initial speed of 21.8 m/s at an angle of 35.0-degrees above the horizontal.
(a) Determine the time of flight of the projectile.
(b) Determine the peak height of the projectile.
(c) Determine the horizontal displacement of the projectile.
67. A projectile is launched horizontally from the top of a 45.2-meter high cliff and lands a distance of 17.6 meters from the base of the cliff. Determine the magnitude of the launch velocity.
68. Two Glenview students stand on the top of their 3.29-meter second-story deck and launch a water balloon from a home-made winger. The balloon is launched upward at a speed of 45.2 m/s and an angle of 39.1 degrees. The balloon lands in a retention pond whose surface is 2.92 meters below grade. Determine the horizontal distance from launch location to landing location.
69. A place kicker kicks a football from 39.6 meters from the goal posts. The kick leaves the ground with a speed of 24.8 m/s at an angle of 49.6 degrees. The goal posts are 3.10-meters high.
(a) Determine the amount by which the kick clears the goal posts.
(b) For this given launch velocity, what is the longest field goal (in yards) which could have been kicked? Assume that the football hits the horizontal cross-bar of the posts and bounces through. Given: 1 meter = 3.28 feet.
70. Sammy Sosa clubs a homerun which sails 421 feet and lands on an apartment balcony located a vertical distance of 59 feet above the level of the ball-bat contact location. An observer times the flight to the balcony to take 3.4 seconds.
(a) Determine the velocity (magnitude and angle) at which the ball leaves the bat.
(b) Determine the speed of theball (in miles/hour) when it lands in the bleachers.
Given: 1 m/s = 2.24 mi/hr; 1 meter = 3.28 feet.
71. An unfortunate accident occurred on the tollway. A driver accidentally passed through a faulty barricade on a bridge (quite unfortunately). and landed in a pile of hay (quite fortunately). Measurements at the accident scene reveal that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. If the driver was in a 65 mi/hr speed zone, then determine the amount by which the driver was exceeding the speed limit at the time of the accident. Assume that the contact with the barricade did not slow the car down. (1 m/s = 2.24 mi/hr)
72. Cupid wishes to shoot an arrow through the open window of a tall building. The window is 32.8 meters above the ground and Cupid stands 63.6 meters from the base of the building. If Cupid aims the arrow at an angle of 51.5 degrees above the horizontal, with what minimum speed must he fire the arrow in order for it to enter the window?
73. In a Physics demonstration, a projectile is launched from a height of 1.23 m above the ground with a speed of 10.6 m/s at an angle of 30.0 degrees above the horizontal.
(a) What horizontal distance from the launch location will the projectile land?
(b) With what speed does the projectile land?
74. A car is parked on a cliff overlooking the sea. The cliff is inclined at an angle of 29 degrees below the horizontal. The negligent driver leaves the car in neutral and it begins rolling from rest towards the cliff's edge with an acceleration 4.5 m/s/s. The car moves a linear distance of 57.2 m to the edge of the cliff before plunging into the ocean below. The cliff is 42.2 m above the sea.
(a) Find the speed (in m/s) of the car the moment it leaves the cliff.
(b) Find the time (in seconds) it takes the car to drop to the water below the edge of the cliff.
(c) Find the position (in meters) of the car relative to the base of the cliff when it lands in the sea.
75. Mike is running to Physics class because he is late at 6.0 m/s. Walking quickly at the same instant from the opposite end of the 20.0 m long hallway, Derek walks toward Mike (and the physics classroom) at –2.5 m/s. When and where do they meet if they don’t stop at the physics classroom? If the classroom is located 7.0 m from Derek’s initial location, who makes it there first?
Draw a position-time graph, using the slopes of the lines, you can roughly determine when the two will pass.
1. Which of the following statements about vectors and scalars are TRUE? List all that apply.
h. A vector is a large quantity and a scalar is a small quantity. i. A scalar quantity has a magnitude and a vector quantity does not. j. A vector quantity is described with a direction and a scalar is not. k. Scalar quantities are path dependent quantities and vector quantities are not. l. A scalar quantity depends only upon the initial and final values of the quantity;
this is not the case for vector quantities. m. The quantity 20 m/s, north is a speed and as such is a scalar quantity. n. The quantity 9.8 m/s/s is an acceleration value and as such is a vector quantity.
Answer: CD
a. FALSE - This would never be the case. Vectors simply are direction-conscious, path-independent quantities which depend solely upon the initial and final state of an object. Vectors are always expressed fully by use of a magnitude and a direction.
b. FALSE - Both scalar and vector quantities have a magnitude or value expressed with a given unit; additionally, a vector quantity requires a direction in order to fully express the quantity.
c. TRUE - Vectors are fully described by magnitude AND direction; scalars are not described with a direction.
d. TRUE - Scalars such as distance would depend upon the path taken from initial to final location. If you run around the track one complete time, your distance will be different than if you take a step forward and a step backwards. The path MATTERS; distance (like all scalars) depends upon it. On the other hand, the displacement (a vector quantity) is the
same for both paths.
e. FALSE - Vectors are the types of quantities which depend only upon initial and final state of the object. For instance, the vector quantity displacement depends only upon the starting and final location.
f. FALSE - This is certainly not a speed quantity; though the unit is appropriate for speed, the statement of the direction is inconsistent with speed as a scalar quantity.
g. FALSE (a rather picky FALSE) - If a direction was included, then this would be an acceleration value. The unit is characteristic of acceleration but the lack of direction is inconsistent with acceleration being a vector quantity.
2. Which of the following statements about distance and/or displacement are TRUE? List all that apply.
i. Distance is a vector quantity and displacement is a scalar quantity. j. A person makes a round-trip journey, finishing where she started. The
displacement for the trip is 0 and the distance is some nonzero value. k. A person starts at position A and finishes at position B. The distance for the trip is
the length of the segment measured from A to B. l. If a person walks in a straight line and never changes direction, then the distance
and the displacement will have exactly the same magnitude. m. The phrase "20 mi, northwest" likely describes the distance for a motion. n. The phrase "20 m, west" likely describes the displacement for a motion. o. The diagram below depicts the path of a person walking to and fro from position
A to B to C to D. The distance for this motion is 100 yds. p. For the same diagram below, the displacement is 50 yds.
Answer: BDF
a. FALSE - Distance is the scalar and displacement is the vector. Know this one!
b. TRUE - Displacement is the change in position of an object. An object which finishes
where it started is not displaced; it is at the same place as it started and as such has a zero displacement. On the other hand, the distance is the amount of ground which is covered. And if it was truly a journey, then there is definitely a distance.
c. FALSE - This would only be the case if the person walk along a beeline path from A to B. But if the person makes a turn and veers left, then right and then ..., then the person has a distance which is greater than the length of the path from A to B. Distance refers to the amount of ground which is covered.
d. TRUE - If a person never changes direction and maintains the same heading away from the initial position, then every step contributes to a change in position in the same original direction. A 1 m step will increase the displacement (read as out of place-ness) by 1 meter and contribute one more meter to the total distance which is walked.
e. FALSE - Distance is a scalar and is ignorant of direction. The "northwest" on this quantity would lead one to believe that this is a displacement (a vector quantity) rather than a distance.
f. TRUE - The unit is an appropriate displacement unit (length units) and the direction is stated. Since there is both magnitude and direction expressed, one would believe that this is likely a displacement.
g. FALSE - The distance from A to B is 35 yds; from B to C is 20 yds; and from C to D is 35 yds. The total distance moved is 90 yds.
h. FALSE (a rather picky FALSE) - Technically, this is not a displacement since displacement is a vector and fully described by both magnitude and direction. The real expression of displacement is 50 yds, left (or west or -)
3. Which of the following statements about velocity and/or speed are TRUE? List all that apply.
k. Velocity is a vector quantity and speed is a scalar quantity. l. Both speed and velocity refer to how fast an object is moving. m. Person X moves from location A to location B in 5 seconds. Person Y moves
between the same two locations in 10 seconds. Person Y is moving with twice the speed as person X.
n. The velocity of an object refers to the rate at which the object's position changes. o. For any given motion, it is possible that an object could move very fast yet have
an abnormally small velocity. p. The phrase "30 mi/hr, west" likely refers to a scalar quantity. q. The average velocity of an object on a round-trip journey would be 0. r. The direction of the velocity vector is dependent upon two factors: the direction
the object is moving and whether the object is speeding up or slowing down.
s. The diagram below depicts the path of a person walking to and from position A to B to C to D. The entire motion takes 8 minutes. The average speed for this motion is approximately 11.3 yds/min.
t. For the same diagram below, the average velocity for this motion is 0 yds/min.
Answer: ADEGI
a. TRUE - Yes! Speed is a scalar and velocity is the vector. Know this one!
b. FALSE - Speed refers to how fast an object is moving; but velocity refers to the rate at which one's motion puts an object away from its original position. A person can move very fast (and thus have a large speed); but if every other step leads in opposite directions, then that person would not have a large velocity.
c. FALSE - Person Y has one-half the speed of Person X. If person Y requires twice the time to do the same distance, then person Y is moving half as fast.
d. TRUE - Yes! That is exactly the definition of velocity - the rate at which velocity changes.
e. TRUE - An Indy Race car driver is a good example of this. Such a driver is obviously moving very fast but by the end of the race the average velocity is essentially 0 m/s.
f. FALSE - The presence of the direction "west" in this expression rules it out as a speed expression. Speed is a scalar quantity and direction is not a part of it.
g. TRUE - For a round trip journey, there is no ultimate change in position. As such, the average velocity is 0 m/t seconds. Regardless of the time, the average velocity will be 0 m/s.
h. FALSE - The direction of the velocity vector depends only upon the direction that the object is moving. A westward moving object has a westward velocity.
i. TRUE - As discussed in #2g, the distance traveled is 90 meters. When divided by time (8 minutes), the average speed is 11.25 yds/min.
j. FALSE - The average velocity would be 0 yds/min only if the person returns to the
initial starting position. In this case, the average velocity is 50 yds/8 min, west (6.25 yds/min, west).
4. Which of the following statements about acceleration are TRUE? List all that apply.
n. Acceleration is a vector quantity. o. Accelerating objects MUST be changing their speed. p. Accelerating objects MUST be changing their velocity. q. Acceleration units include the following; m/s2, mi/hr/sec, cm/s2, km/hr/m. r. The direction of the acceleration vector is dependent upon two factors: the
direction the object is moving and whether the object is speeding up or slowing down.
s. An object which is slowing down has an acceleration. t. An object which is moving at constant speed in a circle has an acceleration. u. Acceleration is the rate at which the velocity changes. v. An object that is accelerating is moving fast. w. An object that is accelerating will eventually (if given enough time) be moving
fast. x. An object that is moving rightward has a rightward acceleration. y. An object that is moving rightward and speeding up has a rightward acceleration. z. An object that is moving upwards and slowing down has an upwards acceleration.
Answer: ACEFGHL (and maybe J)
a. TRUE - Yes it is. Acceleration is direction-conscious.
b. FALSE - Accelerating objects could be changing their speed; but it is also possible that an accelerating object is only changing its direction while maintaining a constant speed. The race car drivers at Indy might fit into this category (at least for certain periods of the race).
c. TRUE - Accelerating object MUST be changing their velocity -either the magnitude or the direction of the velocity.
d. FALSE - The first three sets of units are acceleration units - they include a velocity unit divided by a time unit. The last set of units is a velocity unit divided by a length unit. This is definitely NOT an acceleration.
e. TRUE - This is the case and something important to remember. Consider its application in the last three parts of this question.
f. TRUE - Accelerating objects are either slowing down, speeding up or changing directions.
g. TRUE - To move in a circle is to change one's direction. As such, there is a change in the velocity (not magnitude, but the direction part); this constitutes an acceleration.
h. TRUE - This is the very definition of acceleration. Know this one - its the beginning point of all our thoughts about acceleration.
i. FALSE - Accelerating objects are not necessarily moving fast; they are merely changing how fast they are moving (or the direction they are moving).
j. FALSE - If the accelerating object is slowing down, then it will eventually stop and not reach a fast speed. And if that doesn't convince you, then consider an object that is accelerating by moving in a circle at constant speed forever; it will accelerate the entire time but never being going any faster than at the beginning.
k. FALSE - If an object is moving rightward and slowing down,then it would have a leftward acceleration.
l. TRUE - If an object is speeding up, then the direction of the acceleration vector is in the direction which the object is moving.
m. TRUE - If an object is slowing down, then the acceleration vector is directed opposite the direction of the motion; in this case the acceleration is directed downwards.
5. Which of the following statements about position-time graphs are TRUE? List all that apply.
j. Position-time graphs cannot be used to represent the motion of objects with accelerated motion.
k. The slope on a position-time graph is representative of the acceleration of the object.
l. A straight, diagonal line on a position-time graph is representative of an object with a constant velocity.
m. If an object is at rest, then the position-time graph will be a horizontal line located on the time-axis.
n. Accelerated objects are represented on position-time graphs by curved lines. o. An object with a positive velocity will be represented on a position-time graph by
a line with a positive slope. p. An object with a negative velocity will be represented on a position-time graph by
a line with a negative slope.
q. An object with a positive acceleration will be represented on a position-time graph by a line which curves upwards.
r. An object with a negative acceleration will be represented on a position-time graph by a line which curves downwards.
Answer: CEFG
a. FALSE - Position-time graphs represent accelerated motion by curved lines.
b. FALSE - The slope of a position-time graph is the velocity of the object. Some things in this unit are critical things to remember and internalize; this is one of them.
c. TRUE - A straight diagonal line is a line of constant slope. And if the slope is constant, then so is the velocity.
d. FALSE - Not necessarily true. If the object is at rest, then the line on a p-t graph will indeed be horizontal. However, it will not necessarily be located upon the time axis.
e. FALSE - Accelerating objects (if the acceleration is attributable to a speed change) are represented by lines with changing slope - i.e., curved lines.
f. TRUE - Since slope on a p-t graph represents the velocity, a positive slope will represent a positive velocity.
g. TRUE - Since slope on a p-t graph represents the velocity, a negative slope will represent a negative velocity.
h. FALSE - (This is confusing wording here since we might not all agree on what "curving up" means.) A line that slopes upward and has a curve (perhaps you call that "curving up" as I do) has a positive velocity (due to its positive slope). If the curve is "concave down" (you might say leveling off to a horizontal as time progresses) then the object is slowing down and the acceleration is negative.
i. FALSE - (Once more, there is confusing wording here since we might not all agree on what "curving downwards" means.) A line that slopes downwards and has a curve (perhaps you call that "curving downwards " as I do) has a negative velocity (due to its negative slope). If the curve is "concave up" (you might say leveling off to a horizontal as time progresses) then the object is slowing down and the acceleration is positive.
6. Which of the following statements about velocity-time graphs are TRUE? List all that apply.
j. The slope on a velocity-time graph is representative of the acceleration of the object.
k. The area on a velocity -time graph is representative of the change in position of the object.
l. An accelerated object's motion will be represented by a curved line on a velocity-time graph.
m. Objects with positive acceleration will be represented by upwardly-curved lines on a velocity-time graph.
n. If an object is at rest, then the velocity-time graph will be a line with zero slope. o. A line with zero slope on a velocity-time graph will be representative of an object
which is at rest. p. A line with a negative slope on a velocity-time graph is representative of an object
with negative velocity. q. If an object changes its direction, then the line on the velocity-time graph will
have a changing slope. r. An object which is slowing down is represented by a line on a velocity-time graph
which is moving in the downward direction.
Answer: ABE (and almost D)
a. TRUE - Now this is important! It is the beginning point of much of our discussion of velocity-time graphs. The slope equals the acceleration.
b. TRUE - This is equally important. The area is the displacement.
c. FALSE - An object which has an acceleration will be represented by an line that has a slope. It may or may not curve, but it must have a slope other than zero.
d. FALSE - An object with positive acceleration will have an positive or upward slope on a v-t graph. It does not have to be a curved line. A curved line indicates an object that is accelerating at a changing rate of acceleration.
e. TRUE - An object that is at rest has a 0 velocity and maintains that zero velocity. The permanence of its velocity (not the fact that it is zero) gives the object a zero acceleration. and as such, the line on a v-t graph would have a slope of 0 (i.e., be horizontal).
f. FALSE - A line with zero slope is representative of an object with an acceleration of 0. It could be at rest or it could be moving at a constant velocity.
g. FALSE - A negative slope indicates a negative acceleration. The object could be moving in the positive direction and slowing down (a negative acceleration).
h. FALSE - An object which changes its direction will be represented by a line on a v-t graph that crosses over the time-axis from the + velocity region into the - velocity region.
i. FALSE - An object which is slowing down has a velocity which is approaching 0 m/s. And as such, on a v-t graph, the line must be approaching the v=0 m/s axis.
7. Which of the following statements about free fall and the acceleration of gravity are TRUE? List all that apply.
j. An object that is free-falling is acted upon by the force of gravity alone. k. A falling skydiver which has reached terminal velocity is considered to be in a
state of free fall. l. A ball is thrown upwards and is rising towards its peak. As it rises upwards, it is
NOT considered to be in a state of free fall. m. An object in free fall experiences an acceleration which is independent of the
mass of the object. n. A ball is thrown upwards, rises to its peak and eventually falls back to the original
height. As the ball rises, its acceleration is upwards; as it falls, its acceleration is downwards.
o. A ball is thrown upwards, rises to its peak and eventually falls back to the original height. The speed at which it is launched equals the speed at which it lands. (Assume negligible air resistance.)
p. A very massive object will free fall at the same rate of acceleration as a less massive object.
q. The value of g on Earth is approximately 9.8 m/s2. r. The symbol g stands for the force of gravity.
Answer: ADFGH
a. TRUE - Yes! This is the definition of free fall.
b. FALSE - Skydivers which are falling at terminal velocity are acted upon by large amounts of air resistance. They are experiencing more forces than the force of gravity. As such, they are NOT free-falling.
c. FALSE - Any object - whether rising, falling or moving horizontally and vertically simultaneously - can be in a state of free fall if the only force acting upon it is the force of gravity. Such objects are known as projectiles and often begin their motion while rising upwards.
d. TRUE - The unique feature of free-falling objects is that the mass of the object does not effect the trajectory characteristics. The acceleration, velocity, displacement, etc. is independent of the mass of the object.
e. FALSE - The acceleration of all free-falling objects is directed downwards. A rising object slows down due to the downward gravity force. An upward-moving object which is slowing down is said to have a downwards acceleration.
f. TRUE - If the object is truly in free-fall, then the speed of the object will be the same at all heights - whether its on the upward portion of its trajectory or the downwards portion of its trajectory. For more information, see the Projectiles page at The Physics Classroom.
g. TRUE - The acceleration of free-falling objects (referred to as the acceleration of gravity) is independent of mass. On Earth, the value is 9.8 m/s/s (the direction is down). All objects - very massive and less massive - experience this acceleration value.
h. TRUE - Yes! Know this one!
i. FALSE - Nope. You've never heard (or ever will hear) me call g the force of gravity. g is known as the acceleration of gravity. It might be best to call it the acceleration caused by gravity. When it comes to force of gravity, we have yet another symbol for that - Fgrav. But that's a topic to be discussed in a later unit.
8. If an object has an acceleration of 0 m/s2, then one can be sure that the object is not ____.
a. moving b. changing position c. changing velocity
Answer: C
The object could be moving and could be at rest; however, whether moving or not, it must not have a changing velocity.
9. If car A passes car B, then car A must be ____.
e. accelerating. f. accelerating at a greater rate than car B. g. moving faster than car B and accelerating more than car B. h. moving faster than car B, but not necessarily accelerating.
Answer: D
All that is necessary is that car A has a greater speed (is moving faster). If so, it will eventually catch up and pass car B. Acceleration is not necessary to overcome car B; a car going 60 mi/hr at a constant speed will pass a car going 50 mi/hr at a constant speed. Surely you have witnessed that while driving down Lake Avenue.
10. Which one of the following is NOT consistent with a car which is accelerating?
e. A car is moving with an increasing speed. f. A car is moving with a decreasing speed. g. A car is moving with a high speed. h. A car is changing direction.
Answer: C
An accelerating object must be changing its velocity by either slowing down, speeding up or changing direction. Moving fast merely means that the velocity is high; it says nothing about the acceleration.
11. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. This is evidence that
d. he is accelerating e. he is covering a greater distance in each consecutive second. f. he is moving with a constant speed (on average).
Answer: C
The fullback is moving 10 meters every second. He has a constant speed and thus covers the same distance (10 m) in each consecutive second. He is not accelerating.
12. A fullback is running down the football field in a straight line. He starts at the 0-yard line at 0 seconds. At 1 second, he is on the 10-yard line; at 2 seconds, he is on the 20-yard line; at 3 seconds, he is on the 30-yard line; and at 4 seconds, he is on the 40-yard line. What is the player's acceleration?
Answer: 0 m/s/s
The fullback is moving 10 meters every second. He has a constant speed. He also is running in a straight line, so he is not changing direction. Thus, his acceleration is 0 m/s/s. Only objects with changing velocity have a nonzero acceleration.
13. Olympic gold medalist Michael Johnson runs one time around the track - 400 meters - in 38 seconds. What is his displacement? ___________ What is his average velocity? ___________
Answer: d = 0 m and v = 0 m/s
Michael finishes where he started, so he is not "out of place." His displacement is 0 meters. Since average velocity is displacement over time, his average velocity is also 0 m/s.
14. If an object is moving eastward and slowing down, then the direction of its velocity vector is ____.
a. eastward b. westward c. neither d. not enough info to tell
Answer: A
The direction of the velocity vector is always in the same direction as the direction which the object moves.
15. If an object is moving eastward and slowing down, then the direction of its acceleration vector is ____.
a. eastward b. westward c. neither d. not enough info to tell
Answer: B
If an object is slowing down, then the direction of the acceleration vector is in the opposite direction as the direction which the object moves. (If the object were speeding up, the acceleration would be eastward.)
16. Which one of the following quantities is NOT a vector?
a. 10 mi/hr, east b. 10 mi/hr/sec, west c. 35 m/s, north d. 20 m/s
Answer: D
A vector has both magnitude and direction. Only choice d does not show a direction; it must be a scalar.
17. Which one of the following quantities is NOT a speed?
a. 10 mi/hr b. 10 mi/hr/sec c. 35 m/s d. 20 m/s
Answer: B
You can often tell a quantity by its units. 10 mi/hr/sec is an acceleration since there are
two time units involved. In fact, the units are velocity change units (mi/hr) per time units (seconds). The quantity speed has units of distance/time.
18. Which one of the following statements is NOT true of a free-falling object? An object in a state of free fall ____.
e. falls with a constant speed of -10 m/s. f. falls with a acceleration of -10 m/s/s. g. falls under the sole influence of gravity. h. falls with downward acceleration which has a constant magnitude.
Answer: A
A free-falling object is an object upon which the only force is gravity. As it falls, it accelerates at a rate of approx. 10 m/s/s. This acceleration value is constant during the entire trajectory of the motion. Since this is the case, the speed can not be constant.
19. The average speed of an object which moves 10 kilometers (km) in 30 minutes is ____.
a. 10 km/hr b. 20 km/hr c. 30 km/hr d. more than 30 km/hr
Answer: B
The average speed is distance/time. In this case the distance is 10 km and the time is 0.5 hr (30 minutes). Thus
average speed = (10 km)/(0.5 hr) = 20 km/hr
20. What is the acceleration of a car that maintains a constant velocity of 55 mi/hr for 10 seconds?
a. 0 b. 5.5 mi /hr/s c. 5.5 mi /s/s d. 550 mi/hr/s
Answer: A
If the velocity is constant, then there is no acceleration. That is, the value of the acceleration is 0.
21. As an object freely falls, its ____.
a. speed increases b. acceleration increases
c. both of these d. none of these
Answer: A
As an object falls, it accelerates; this means that the speed will be changing. While falling, the speed increases by 10 m/s every second. The acceleration is a constant value of 10 m/s/s; thus, choice b should not be chosen.
22. A speedometer is placed upon a free-falling object in order to measure its instantaneous speed during the course of its fall. Its speed reading (neglecting air resistance) would increase each second by about ____.
a. 5 m/s. b. 10 m/s. c. 15 m/s. d. a variable amount. e. depends on its initial speed.
Answer: B
The acceleration of gravity is approximately 10 m/s/s. Acceleration represents the rate at which the velocity changes - in this case, the velocity changes by 10 m/s every second. So the speed will increase by the amount of 10 m/s every second.
23. Ten seconds after being dropped from rest, a free-falling object will be moving with a speed of about ____.
a. 10 m/s. b. 50 m/s. c. 100 m/s. d. more than 100 m/s.
Answer: C
Since the speed of a free-falling object increases by 10 m/s every second, the speed after ten of these seconds will be 100 m/s. You could use the kinematic equation
vf = vi + a*t
where vi=0 m/s and a = -10 m/s/s and t=10 s
24. A baseball pitcher delivers a fast ball. During the throw, the speed of the ball increases from 0 to 30 m/s over a time of 0.10 seconds. The average acceleration of the baseball is ____ m/s2.
a. 3 b. 30 c. 300 d. 3000 e. none of theseAnswer: C
Acceleration is velocity change over time. In this problem, the velocity change is +30 m/s and the time is 0.1 s. Thus,
a = (+30 m/s)/(0.1 s) = 300 m/s/s.
25. On takeoff, a rocket accelerates from rest at a rate of 50 m/s2 for 1 minute. The rocket's speed after this minute of steady acceleration will be ____. m/s
a. 50 b. 500 c. 3000 d. 3600 e. none of these
Answer: C
Use the equation
vf = vi + a*t
vf = 0 + (50 m/s/s)*(60 s) = 3000 m/s
(Note that the unit on time must be the same as the time units for which the acceleration is given.)
26. When a rock is dropped, it will accelerate downward at a rate of 9.8 m/s2. If the same rock is thrown downward (instead of being dropped from rest), it acceleration will be ____. (Ignore air resistance effects.)
a. less than 9.8 m/s2 b. 9.8 m/s2 c. more than 9.8 m/s2
Answer: B
Whether rising or falling, if the sole force acting upon the object is gravity, then the acceleration is 9.8 m/s/s (often approximated as 10 m/s/s).
27. Consider drops of water that leak from a dripping faucet at a constant rate. As the drops fall they ____.
a. get closer together. b. get farther apart.
c. remain at a relatively fixed distance from one another.
Answer: B
Since the drops of water are falling (and probably free-falling), they should be getting farther apart as they fall. This is because the free-falling drops are accelerating and thus gaining speed.
28. Renatta Oyle is again found driving her '86 Yugo down Lake Avenue, leaving the following trail of oil drops on the pavement.
If her car is moving from right to left, then ...
e. her velocity has a rightward direction and her acceleration has a rightward direction.
f. her velocity has a rightward direction and her acceleration has a leftward direction.
g. her velocity has a leftward direction and her acceleration has a rightward direction.
h. her velocity has a leftward direction and her acceleration has a leftward direction.
Answer: D
The car is heading leftward and the velocity is always in the same direction as the direction which the object moves. Since the car is speeding up, the acceleration is leftward. Whenever an object speeds up, its acceleration is in the same direction which the object moves. Whenever an object slows down, its acceleration is in the opposite direction which the object moves.
29. On the diagrams below, construct a dot diagram representing the motion of an object with a ... .
f. constant rightward velocity g. rightward velocity and a rightward acceleration h. rightward velocity and a leftward acceleration i. rightward velocity, first slow and constant, and then accelerating to a high speed j. rightward velocity, first decelerating from a high speed to a rest position, then
maintaining the rest position, and finally accelerating at a lower rate than the initial deceleration.
Answer: See diagram above; explanations are given below.
a. A constant velocity is depicted by dots which are spaced the same distance apart.
b. An object which is accelerating in the same direction as the velocity (both are rightward in this case) is speeding up. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced further and further apart to indicate that the object is speeding up.
c. An object that moves rightward (i.e., rightward velocity) and accelerating leftwards must be slowing down. And so as you trace your eye from left to right across the diagram (rightward), the dots should be spaced closer and closer together to indicate that the object is slowing down.
d. The dots begin by being equally spaced (constant speed) and close together (slow); then the spacing between dots become gradually further and further apart to indicated the speeding up nature of the motion.
e. The object is moving rightwards so you will need to trace your eye from left to right across the diagram (rightward). As you do you will note that the dots become closer and closer to indicate that the object is slowing down. Then the dots are piled onto the same location to indicate that the object is at rest. Finally, the dots are spread further and further apart; the rate at which the distance between dots is very gradual - consistent with the statement that the rate of acceleration is less than the original rate of deceleration.
30. On a dot diagram, how does the motion of an object moving to the right and slowing down differ from an object moving to the left and speeding up? Explain.
Answer: There is no difference!
The two diagrams would be indistinguishable from each other as indicated by the following two diagrams.
There are two ways to accelerate leftward; and regardless of how it happens, the oil drop trace would be identical. A leftward acceleration is a leftward acceleration.
31. On the position-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.)
h. at rest. i. moving in the positive direction with constant speed j. moving in the negative direction and speeding up k. moving in the positive direction and slowing down l. moving in the positive direction at a constant speed (slow) and then later fast at
constant speed m. moving with a negative velocity and a negative acceleration n. moving with a negative velocity and a positive acceleration
Answer: See diagram above; explanations are given below.
The governing principle in this problem is that the slope of a position-time graph is equal to the velocity of the object.
a. An object at rest (v = 0 m/s) is represented by a line with 0 slope (horizontal line).
b. An object with a constant, positive velocity is represented by a straight line (constant slope) which slopes upward (positive slope).
c. An object moving in the - direction and speeding up is represented by a line which slopes downward (- slope) and increases its steepness (increasing slope).
d. An object moving in the + direction and slowing down is represented by a line which slopes upward (+ slope) and increases its steepness (increasing slope).
e. An object moving in the positive direction at constant speed would be represented by a straight diagonal line (constant speed) which slopes upward (+ velocity). So in this case, there will be two straight diagonal lines; the second line will slope more than the first line.
f. An object moving in the negative direction with negative acceleration is speeding up (since the a vector is in the same direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and increases its slope over time (speeding up).
g. An object moving in the negative direction with positive acceleration is slowing down (since the a vector is in the opposite direction as the motion). So on a p-t graph, this object will be represented by a line which slopes downwards (- velocity) and levels off or becomes more horizontal over time (slowing down).
32. On the velocity-time graph below, sketch a plot representing the motion of an object which is ... . Label each line with the corresponding letter (e.g., "a", "b", "c", etc.)
i. at rest j. moving in the positive direction at constant speed k. moving in the negative direction from slow to fast
l. moving in the negative direction from fast to slow m. moving with a positive velocity and a positive acceleration n. moving with a positive velocity and a negative acceleration o. moving with a positive velocity at constant speed and then decelerating to a rest
position p. moving in the positive direction while slowing down, changing directions and
moving in the negative directions while speeding up
Answer: See diagram above; explanations are given below.
The governing principle in this problem is that the slope of a velocity-time graph is equal to the acceleration of the object. Furthermore, a negative velocity would be a line plotted in the negative region of the graph; a positive velocity would be a line plotted in the positive region of the graph.
a. An object at rest (v = 0 m/s) is represented by a line located on the time axis (where v = 0 m/s).
b. An object moving in the positive direction at constant speed will be represented on a v-t
graph by a horizontal line (slope = a = 0 m/s/s) positioned in the + velocity region.
c. An object moving in the negative direction and speeding up will be represented on a v-t graph by a sloped line located in the - velocity region. Since such an object has a - acceleration, the line will slope downwards (- acceleration).
d. An object moving in the negative direction and slowing down will be represented on a v-t graph by a sloped line located in the - velocity region. Such an object has a positive acceleration (since it is slowing down, the a vector will be in the opposite direction of the motion). The + acceleration would be consistent with a line that slopes upwards.
e. An object moving with a + velocity and a + acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The + acceleration would be consistent with a line that slopes upwards.
f. An object moving with a + velocity and a - acceleration would be represented on a v-t graph by a sloped line located in the + velocity region. The - acceleration would be consistent with a line that slopes downwards.
g. An object moving with a constant + velocity would be represented on a v-t graph by a horizontal line (slope = a = 0 m/s/s) located in the + velocity region. The slowing down to rest portion of the graph would be represented by a line which slopes downwards (- acceleration) towards the time axis.
The velocity-time graph below depicts the motion of an automobile as it moves through Glenview during rush hour traffic. Use the graph to answer questions #35 - #39.
33. Determine the displacement of the automobile during the following intervals of time. PSYW
Answer: See diagram above and calculations below.
Use area calculations to determine the displacement of the object.
t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s
Area of the pink triangle:
A = 0.5*b*h
A = 0.5*(5 s)*(10 m/s)
A = 25 m
Area of the green rectangle:
A = b*h
A = (10 s)*(10 m/s)
A = 100 m
Area of the blue triangle
plus purple rectangle:
A = 0.5*b1*h1 + b2*h2
A = 0.5*(5 s)*(5 m/s) + (5 s)*(5 m/s)
A = 12.5 m + 25 m
A = 37.5 m
34. Determine the velocity of the automobile at the following instant(s) in time.
Answer: See calculations below.
It is a velocity graph; so merely read the velocity values off the graph.
t = 3 s t = 8 s t = 17 s
v = 6.0 m/s
v = 10.0 m/s
v = 8.0 m/s
35. Determine the acceleration of the automobile during the following intervals of time.
Answer: See calculations below.
Do slope calculations to determine the acceleration. Slope is rise/run.
t = 0 s - 5 s t = 5 s - 15 s t = 15 s - 20 s
a = slope = rise/run
a = slope = rise/run
a = slope = rise/run
a = (10 m/s)/(5 s)
a = 2.0 m/s/s
a = (0 m/s)/(10 s)
a = 0.0 m/s/s
a = (-5 m/s)/(5 s)
a = -1.0 m/s/s
36. Using complete sentences and the language of physics, describe the motion of the automobile during the entire 20 seconds. Explicitly describe any changes in speed or direction which might occur; identify intervals of time for which the automobile is at rest, the automobile is moving with constant speed, or the automobile is accelerating.
Answer:
During the first 5 seconds, the auto is moving with a + velocity and a + acceleration, increasing its speed from slow to fast. From 5 s to 15 s, the auto is moving with a constant speed and a zero acceleration. From 15 s to 20 s, the auto is moving with a +velocity and a - acceleration, decreasing its speed from fast to slow.
37. Supposing the automobile has an oil leak, demonstrate your understanding of its motion by drawing an oil drop diagram for the 20 seconds of motion. Divide the diagram into three distinct time intervals (0-5 seconds, 5-15 seconds, 15-20 seconds).
Answer: See Diagram and explanation below.
In the first five seconds, the object is speeding up so the spacing between dots will gradually increase. In the next ten seconds (5-15 s) the object is maintaining a constant velocity so the spacing between dots remains the same. In the last 5 seconds (15-20 s) the object is slowing down so the spacing between dots is gradually decreasing.
38. For the plots below, determine the velocity of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. at 13 seconds
Answer: See explanations and calculations below.
For a position-time graph, the velocity is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed.
a. Pick the two points: (0 s, 20 m) and (5 s, 10 m)
slope = rise/run = (-10 m)/(5 s) = -2.0 m/s
b. Pick the two points: (5 s, 2 m) and (10 s, 6 m)
slope = rise/run = (4 m)/(5 s) = +0.80 m/s
c. From 0 to 20 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (0 s, 10 m) and (20 s, 50 m)
slope = rise/run = (40 m)/(20 s) = +2.0 m/s
39. For the plots below, determine the acceleration of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. at 13 seconds
Answer: See explanations and calculations below.
For a velocity-time graph, the acceleration is determined from a slope calculation. So in each case, two points must be picked and slope calculation must be performed.
a. Pick the two points: (0 s, 5 m/s) and (5 s, 30 m/s)
slope = rise/run = (25 m/s)/(5 s) = +5.0 m/s2
b. Pick the two points: (5 s, 25 m/s) and (10 s, 10 m/s)
slope = rise/run = (-15 m/s)/(5 s) = -3.0 m/s2
c. From 10 to 15 s, the slope is a constant value. So determining the slope at 13 seconds is as simple as merely determining the slope of the line choosing any two points. So just pick the two points: (10 s, 15 m/s) and (15 s, 0 m/s)
slope = rise/run = (-15 m/s)/(5 s) = -3.0 m/s2
40. For the plots below, determine the displacement of the object ... .
a. from 0-5 seconds b. from 5-10 seconds c. from 0 -15 seconds
Answer: See explanations and calculations below.
For velocity-time graphs, the displacement of an object is found by computing the area between the line and the time axis. The shape is typically a rectangle (area = base*height), a triangle (area = 0.5*base*height) or a trapezoid (which can typically be transformed into a rectangle and a triangle or the area can be computed as 0.5*(h1 + h2)*base).
a. Area = 0.5*base*height = 0.5*(5 s)*(20 m/s) = 50 m
b. Area = A1 + A2 (see diagram) = (5 s)*(5 m/s) + 0.5*(5 s)*(10 m/s) = 25 m + 25 m = 50 m
41. Determine the acceleration (in m/s2) of an object which ... .
e. moves in a straight line with a constant speed of 20 m/s for 12 seconds f. changes its velocity from 12.1 m/s to 23.5 m/s in 7.81 seconds g. changes its velocity from 0 mi/hr to 60 mi/hr in 4.2 seconds h. accelerates from 33.4 m/s to 18.9 m/s over a distance of 109 m
Answer: See answers, explanations and calculations below.
a. If the speed and direction of an object is constant, then the acceleration is 0 m/s2.
b. The acceleration is the velocity change per time ratio:
a = (Velocity Change)/t = (23.5 m/s - 12.1 m/s) / (7.81 s) = 1.46 m/s2.
c. The acceleration is the velocity change per time ratio:
d. The acceleration value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 33.4 m/s; vf = 18.9 m/s; and d = 109 m. Using the equation vf
42. Determine the magnitude of the displacement (in meters) of an object which ... .
f. moves from Hither to Yon (with an average speed of 28 m/s) and then back to Hither (with an average speed of 28 m/s) if both the forward and the return trip take 46 minutes each.
g. moves at a constant speed of 8.3 m/s in a straight line for 15 seconds. h. decelerates at a rate of -4.35 m/s/s from a speed of 38.1 m/s to a speed of 17.6 m/s
i. accelerates from rest at a rate of 3.67 m/s2 for 12.1 seconds j. is moving at 12.2 m/s and then accelerates at a rate of +1.88 m/s2 for 17 seconds
Answer: See answers, explanations and calculations below.
a. Since this is a round-trip journey, the overall displacement is 0 m.
b. Since the velocity is constant, the displacement can be found by multiplying the velocity by the time.
d = v*t = (8.3 m/s) * (15 s) = ~125 m
c. A displacement value can also be calculated using kinematic equations if three other kinematic quantities are known. In this case, the know information is: vo = 38.1 m/s; vf = 17.6 m/s; and a = -4.35 m/s/s. Using the equation vf
d. A displacement value can be calculated using other kinematic equations if a different set of kinematic quantities is known. Here we know that: vo = 0.0 m/s; t = 12.1 s; and a = 3.67 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.
d = (0 m/s)*(12.1 s) + 0.5*(3.67 m/s/s)*(12.1 s)2 = 269 m.
e. Here the displacement value is calculated using the same kinematic equation. We know that: vo = 12.2 m/s; t = 17.0 s; and a = 1.88 m/s/s. Using the equation d = vo* t + 0.5*a*t2, the displacement can be computed.
d = (12.2 m/s)*(17.0 s) + 0.5*(1.88 m/s/s)*(17.0 s)2 = 479 m.
The tortoise, moving at a constant speed, will cover the 1200 m in a time of:
ttortoise = d/vtortoise = (1200 m) / (0.05 m/s) = 24000 s = 6.667 hrs
The hare will sleep for 6.5 hrs (23400 s) before starting, and so will have only 0.167 hrs (600 s) to accelerate to the finish line. So the acceleration of the hare can be determine
using a kinematic equation. The known information about the hare's motion is: t = 600 s; d = 1200 m; vo = 0 m/s. The best equation is d = vo* t + 0.5*a*t2. The vo* t term cancels and the equation can be algebraically rearranged and solved for a:
a = 2*d / t2 = 2 * (1200 m) / (600 s)2 = 0. 667 m/s2.
44. A Gold Car moving at 12 m/s/s passes a Green Car while the Green Car is at rest at a stoplight. The Green Car immediately accelerates at a rate of +1.8 m/s for 11 seconds. After how much time (relative to the initial starting time) must the Green Car drive before catching up with the Gold Car.
Answer: 14 s
(As mentioned in the previous problem ...) Like a lot of physics word problems, there is more than one path to the final answer. In all such problems, the solution involves thought and good problem-solving strategies (draw a picture, list what you know, list pertinent equations, etc.).
Here the gold car travels with a constant speed for a time of t seconds (where t is the total time of travel for both cars). The distance traveled by the gold car is given by the kinematic equation d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as
d = vo* t + 0.5*a*t2 = (12 m/s)*t, or
dgold = 12* t
For the green car, there is an accelerated period and then a constant speed period. The distance traveled during the accelerated period (d1green) is found from the same kinematic equation. For the green car, the first term cancels and the distance is
d1green = vo* t + 0.5*a*t2 = 0.5*(1.8 m/s2)*(11 s)2, or
d1green = 108.9 m
Once the green car has accelerated for 11 seconds, it maintains a constant speed for the remaining time, given by the expression t - 11 s. The speed at which the green car is traveling during this time can be computed using the equation:
vfgreen = vo + a*t = (1.8 m/s2) *(11 s) = 19.8 m/s
The distance traveled by the green car during this constant speed portion of its motion (d2green) can be computed using the kinematic equation. d = vo* t + 0.5*a*t2. The second term cancels and the distance can be expressed as
d2green = vo* t + 0.5*a*t2 = (19.8 m/s) * (t - 11 s) = 19.8*t - 217.8, or
d2green = 19.8*t - 217.8
So the total distance traveled by the green car is given by the expression:
When the green car catches up to the gold car their distance traveled will be the same. So the time t can be determined by setting the two expressions for distance equal to each other and solving for t.
12* t = 19.8*t - 108.9
108.9 = 7.8*t
t = (108.9) / (7.8)
t = 13.96 s = ~14 s
45. Ima Rilla Saari is cruising at 28 m/s down Lake Avenue and through the forest preserve. She notices a deer jump into the road at a location 62 m in front of her. Ima first reacts to the event, then slams on her brakes and decelerates at -8.1 m/s2, and ultimately stops a picometer in front of the frozen deer. What is Ima's reaction time? (i.e., how long did it take Ima to react to the event prior to decelerating?)
Answer: 0.49 s
Ima's total distance traveled (62 m) can be broken into two segments - a reaction distance (drxn) and a braking distance (dbraking). The reaction distance is the distance Ima moves prior to braking; she will move at constant speed during this time of trxn. The braking distance is the distance which Ima moves when her foot is on the brake and she decelerates from 28 m/s to 0 m/s. The braking distance can be computed first using the following kinematic equation: vf
2 = vo2 + 2*a*d. The known information for this braking period is: vo = 28 m/s;
vf = 0 m/s; and a = -8.1 m/s/s. The substitutions and solution are shown below.
Since Ima's car requires 48.40 m to brake, she can travel a maximum of 13.6 m during the reaction period. The relationship between reaction time, speed and reaction distance is given by the equation
drxn = v * trxn
Substituting 13.6 m for drxn and 28 m/s for v, the reaction time can be computed:
trxn = (13.6 m) / (28 m/s) = 0.486 s
46. A two-stage rocket accelerates from rest at +3.57 m/s2 for 6.82 seconds. It then accelerates at +2.98 m/s2 for another 5.9 seconds. After the second stage, it enters into a state of free fall. Determine:
e. the maximum speed f. the maximum altitude g. the height of the rocket after 20.0 seconds h. the total time the rocket is in the air (assuming it is launched from the ground)
Answer: See answers and explanations below.
This problem can be approached by either the use of a velocity-time graph or the use of kinematic equations (or a combination of each). Whatever the approach, it is imperative to break the multistage motion up into its three different acceleration periods. The use of kinematic equations is only appropriate for constant acceleration periods. For this reason, the complex motion must be broken up into time periods during which the acceleration is constant. These three time periods can be seen on the velocity-time graph by three lines of distinctly different slope. The diagram at the right provides a depiction of the motion; strategic points are labeled. These points will be referred to in the solutions below. The velocity-time plot below will be used throughout the solution; note that the same strategic points are labeled on the plot.
a. The maximum speed occurs after the second stage or acceleration period (point C). After this time, the upward-moving rocket begins to slow down as gravity becomes the sole force acting upon it. To determine this speed (vc), the kinematic equation vf = vo + a*t will be used twice - once for each acceleration period.
First Stage: vB = vA + a*t = 0 m/s + (3.57 m/s/s) * (6.82 s) = 24.3 m/s
Second Stage: vC = vB + a*t = 24.3 m/s + (2.98 m/s/s) * (5.90 s) = 41.9 m/s
b. The maximum altitude occurs at point D, sometime after the second stage has ceased and the rocket finally runs out of steam. The velocity at this point is 0 m/s (it is at the peak of its trajectory). The altitude at this point is the cumulative distance traveled from t = 0 s to t = tD. This distance is the distance for the first stage, the second stage and the deceleration period (C to D). These distances correspond to the area on the v-t graph; they are labeled A1, A2, and A3 on the graph. They are calculated and summed below.
It will be necessary to know the time from point C to point D in order to determine A3. This time can be determined using the kinematic equation vf = vo + a*t for which vf = 0 m/s and vo = 41.9 m/s and a = -9.8 m/s/s.
vD = vC + a*t
0 m/s = 41.9 m/s + (-9.8 m/s/s) * t
t = 4.28 s
Now A3 can be determined using the v-t graph. The area is a triangle and is calculated as
The maximum altitude is the sum of the three distances (areas)
Max. altitude = 82.86 m + 196.65 m + 89.57 m = ~369 m
c. When the rocket reaches point D, the time is 17.0 seconds. The altitude at 20.0 seconds will be the 369 meters risen above the launch pad from point A to point D minus the distance fallen from the peak from 17.0 to 20.0 seconds. This distance would be represented by a negative area on the velocity-time graph. The area is a triangle and can be computed if the velocity at 20 seconds is known. It can be calculated using a kinematic equation and then used to determine the area of a triangle. Alternatively, a kinematic equation can be used to determine the distance fallen during these 3.0 seconds. The work is shown below:
d = vo* t + 0.5*a*t2 = 0.5 * (-9.8 m/s/s) * (3.0 s)2 = 44.1 m
The altitude at 20 seconds is therefore the ~369 m risen in the first 17 seconds minus the ~44 m fallen in the next 3 seconds. The answer is ~325 m.
d. The rocket rises 369 m in the first 17.0 seconds. In the time subsequent of this, the rocket must fall 369 meters. The time to fall 369 m can be found from the same kinematic equation used in part c.
d = vo* t + 0.5*a*t2
-369 m = 0.5 * (-9.8 m/s/s) * t2
t = 8.68 seconds
This time can be added onto the 17.0 seconds to determine the time at which the rocket lands: ~25.7 seconds.
47. In a 200.0-m relay race (each leg of the race is 50 m long), one swimmer has a 0.45 second lead and is swimming at a constant speed of 3.9 m/s towards the opposite end of the pool. What minimum speed must the second swimmer have in order to catch up with the first swimmer by the end of the pool?
Answer: 4.0 m/s
Both swimmers swim the same distance (50 m) at constant speed. Swimmer A (who was just arbitrarily named) gets a 0.45 second start. So swimmer B must travel faster in order to finish the race in less time than swimmer A. First, the time required of swimmer A to complete the 50 m at 3.9 m/s can be computed. The time is
tA = d/vA = (50 m) / (3.9 m/s) = 12.82 s
Thus, swimmer B must finish the same 50 m in 12.37 seconds (12.82s - 0.45 s). So the speed of swimmer B can be computed as
vB = d/tB = (50 m) / (12.37 s) = 4.04 m/s
48. A drag racer accelerates from rest at an average rate of +13.2 m/s2 for a distance of 100.0 m. The driver coasts for 0.50 seconds and then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the end of the track?
Answer: -24 m/s/s
This problem can be approached by first determining the distance over which the dragster decelerates. This distance will be less than 80 meters by an amount equal to the distance which the dragster coasts after crossing the finish line. See diagram.
The distance traveled by the dragster prior to braking is 100 m plus the coasting distance. The coasting distance can be determined if the speed of the dragster at the end of 100 m is determined. So first, a kinematic equation will be used to determine the speed and then the coasting distance will be computed.
Using the equation vf2 = vo
2 + 2*a*d, the speed after 100 m can be determined. This substitution and solution is shown below.
vf2 = vo
2 + 2*a*d = 2*(13.2 m/s/s)*(100 m) = 2640 m2/s2
vf = 51.38 m/s
Coasting at 51.38 m/s for 0.5 s will lead to a distance traveled of 25.69 m.
Once the coasting period is over, there is a short distance left to decelerate to a stop. This distance is
180 m - 100 m - 25.69 m = 54.31 m
Now the same kinematic equation can be used to determine the deceleration rate during the last 54.31 m of the track. Known information is: vo = 51.38 m/s; vf = 0 m/s; and d = 54.31 m. Using the equation vf
2 = vo2 + 2*a*d, the acceleration can be computed.
49. Which of the following statements are true of scalars and vectors? List all that are TRUE.
e. A vector quantity always has a direction associated with it. f. A scalar quantity can bave a direction associated with it. g. Vectors can be added together; scalar quantites cannot. h. Vectors can be represented by an arrow on a scaled diagram; the length of the
arrow represents the vector's magnitude and the direction it points represents the vector's direction.
Answer: AD
a. TRUE - Vectors are defined as quantities which are fully described by both their magnitude and direction. By definition, a vector has a direction associated with it. If it didn't, then it would NOT be a vector.
b. FALSE - Scalars are defined as quantities which are fully described by their magnitude alone. Scalars have no regard for direction and it is meaningless to associate a direction with such a quantity. If a quantity did have a direction associated with it, then that quantity would not be a vector.
c. FALSE - Both vectors and scalars can be added together. The rules for adding vectors together are unique to vectors and cannot be used when adding scalars together. The direction of a vector must be considered when adding two vectors together. Direction is of no importance when adding scalars.
d. TRUE - This is exactly the case and exactly what is done throughout the unit.
50. Which of the following quantities are vectors? Include all that apply.
g. distance traveled h. displacement i. average speed j. average velocity k. instantaneous velocity l. acceleration
Answer: BDEF
Of the five kinematic quantities listed here (distance, displacement, speed, velocity and acceleration), three of them are vectors. Displacement, velocity (both average and instantaneous), and acceleration all require the mention of a direction in order to fully describe the quantity.
51. Numerical values and directions are stated for a variety of quantities. Which of these statements represent a vector description? Include all that apply.
i. 20 meters, west j. 9.8 m/s/s k. 35 mi/hr, south l. 16 years old m. 60 minutes n. 3.5 m/s/s, south o. -3.5 m/s/s p. +20 degrees C
Answer: ACFG
Expressions of vector quantities would include a magnitude (number, value, etc.) and a direction. The direction could be described as being north, south, east, west or left, right, up, down. On occasion, a "+" or "-" is used to describe the direction. Since mathematical computations on calculators do not fare well with the typing of "south," a - sign is often
substituted for a given direction. In the case of g, the units indicate an acceleration quantity. The "-" sign indicates a direction. One must be careful in assuming that a "+" or "-" sign is a sure sign of a quantity being a direction for other non-vector quantities can use such signs as well (as is the case in h).
52. Which of the following statements are true of the time of flight for a projectile? List all that apply.
f. The time that a projectile is in the air is dependent upon the horizontal component of the initial velocity.
g. The time that a projectile is in the air is dependent upon the vertical component of the initial velocity.
h. For a projectile which lands at the same height that it is projected from, the time to rise to the peak is equal to the time to fall from its peak to the original height.
i. For the same upward launch angles, projectiles will stay in the air longer if the initial velocity is increased.
j. Assume that a kicked ball in football is a projectile. If the ball takes 3 seconds to rise to the peak of its trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground.
Answer: BCD
a. FALSE - The time for a projectile to rise vertically to its peak (and subsequently fall back to the ground) is dependent upon the initial vertical velocity. Alteration in the horizontal velocity will only cause the projectile to have a greater horizontal displacement (x).
b. TRUE - Absolutely true. Projectiles with a greater vertical component of initial velocity will be in the air for longer amount of times (assuming that the direction of viy is upward). An alteration in the viy value will alter the time of flight of the projectile, regardless of the direction of viy.
c. TRUE - For projectiles launched at upward angles and landing at the original height, the time to the rise to the peak equals the time to fall from the peak. If it takes 3 seconds to rise upward, it will take 3 seconds to fall.
d. TRUE - For a constant launch angle, an increase in the initial velocity (vi) will increase the vertical velocity (viy). This results in an increased time for the projectile to decelerate to 0 m/s as it rises towards its peak. So the projectile takes longer to get to the peak, longer to fall from the peak and overall is in the air for a longer time.
e. FALSE - Close, but very false. If it takes 3 seconds to rise to the peak, then it takes 3 seconds to fall from the peak; The 6 seconds is the total time of flight of the projectile.
53. An object is undergoing free fall motion. As it falls, the object's ____.
a. speed increases b. acceleration increases
c. both of these d. none of these
Answer: A
As an object free-falls, its velocity (and also its speed) changes by approximately 10 m/s every second. This means that the acceleration is a constant value of 10 m/s/s. An object has a changing speed (or velocity) and a constant acceleration if the speed changes by the same amount (a "constant amount") in each consecutive second of its motion.
54. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is _______.
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
Answer: B
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. At its peak, its vertical velocity becomes 0 m/s. At this instant in time, the velocity is entirely horizontal; there is no vertical component to the velocity.
55. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its acceleration is _______. (Neglect the effects of air resistance.)
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
Answer: A
As a projectile rises towards its peak, its horizontal velocity remains constant while its
vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. At the peak and everywhere throughout the trajectory, there is a vertical (downward) acceleration. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.
56. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, the net force acting upon it is _______. (Neglect the effects of air resistance.)
a. entirely vertical b. entirely horizontal
c. both vertical and horizontal d. not enough information given to know.
Answer: A
A projectile is an object upon which the only force is gravity. Since no other forces act upon the object, the net force would be downward.
57. At what point in its path is the horizontal component of the velocity (vx) of a projectile the smallest?
a. The instant it is thrown. b. Halfway to the top. c. At the top.
d. As it nears the top. e. It is the same throughout the path.
Answer: E
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. Having a constant horizontal velocity, there is no point along the trajectory where the vx value is smaller than at other points.
58. At what point in its path is the vertical component of the velocity (vy) of a projectile the smallest?
a. The instant it is thrown. b. Halfway to the top. c. At the top.
d. As it nears the top. e. It is the same throughout the path.
Answer: C
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. During the upward portion of its trajectory, the vy continuously decreases until it becomes 0 m/s at the peak. Thus, the vy is as small as it will ever be when it is at the peak of the trajectory.
59. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity ___.
a. decreases b. remains constant c. increases
Answer: B
Once the ball leaves the table's edge, it becomes a projectile. As it falls, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.
60. A bullet is fired horizontally and hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it would have hit the ground in ____. (Assume no air resistance)
a. less than 0.5 s. b. more than 0.5 s. c. 0.5 s.
Answer: C
Once the bullet leaves the muzzle, it becomes a projectile (assuming no air resistance). As it falls, its horizontal velocity remains constant while its vertical velocity decreases. The force of gravity acts upon the bullet to cause its downward acceleration. The motion of the bullet in the downward direction is independent of the motion in the horizontal direction. That is to say, any alteration in a horizontal aspect of its motion will not effect the motion in the vertical direction. The time to fall vertically to the ground is not effected by the
horizontal speed of the projectile. It would still take 0.5 seconds to fall to the ground from this height regardless of the horizontal speed.
61. A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what other projection angle would produce the same down-range distance?
a. 30 degrees. b. 45 degrees. c. 50 degrees. d. 75 degrees e. 90 degrees.
Answer: D
For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Any two launch angles which are separated from 45 degrees by the same amount (for example, 40 degrees and 50 degrees, 30 degrees and 60 degrees and 15 degrees and 75 degrees) will provide the same horizontal displacement.
62. Two projectiles are fired at equal speeds but different angles. One is fired at angle of 30 degrees and the other at 60 degrees. The projectile to hit the ground first will be the one fired at (neglect air resistance) ____.
a. 30 degrees b. 60 degrees c. both hit at the same time
Answer: A
For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Launch angles greater than 45 degrees result in large vertical components of velocity; these stay in the air longer and rise to higher heights. Launch angles less than 45 degrees result in small vertical components of velocity; these do not rise as high and end up falling to the ground in shorter times.
63. Consider the trajectory diagram shown below for a non-horizontally launched projectile. On the diagram, draw vector arrows representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first position.)
Answer: See diagram above.
A projectile has a constant horizontal velocity and a changing vertical velocity. The changing vertical component of velocity is consistent with the fact that there is a vertical acceleration. As the projectile rises towards the peak of its trajectory, the vertical velocity decreases until it reaches 0 m/s at the very peak of the trajectory. As the projectile rises towards the peak of its trajectory, the vertical velocity increases. The vertical velocity upon falling is of the same magnitude and the opposite direction as any corresponding location of the same height during the rising motion. The horizontal velocity remains constant throughout the trajectory. These principles are shown in the diagram above.
64. Consider the diagram below. The blue path represents the trajectory of a projectile dropped from rest from the top of the path. (Each ball location represents the location after a 1-second interval of time.) The red path represents the trajectory of the same ball thrown horizontally in the absence of gravity. Construct a third path accurately showing the trajectory of a projectile launched with the same horizontal speed as the red ball.
Show the location during the first four seconds of motion. Finally, the (x, y) coordinate position of location 1 is (0 m, -5 m) and the (x, y) coordinate position of location 2 is (15 m, 0 m). Determine the (x, y) coordinate positions of the four locations in your trajectory. (Assume g ~10 m/s/s.)
Answer: See diagram above.
A projectile such as this has both a horizontal and vertical component of motion. These two components of motion are independent of each other. The strictly vertical motion of the ball (the blue path above) depicts the vertical displacement at 1-second intervals. If a horizontal motion is imparted to the ball, its vertical displacement will not be effected since these two components are independent of each other. The strictly horizontal motion of the ball (the red path above) portrays the motion of the ball in the absence of gravity. The presence of gravity would cause the ball to accelerate downward. This vertical motion does not effect the horizontal motion since these two components are independent of each other. The green path represents the motion of the projectile when gravity acts upon it and a horizontal motion is imparted to it. The horizontal location at each 1-second instant in time is the same position as it was in the red path above it. And the vertical location at
each 1-second instant in time is the same position as it was in the blue path to the left of it.
The actual coordinate positions can be determined using the kinematic equations and the given time. For the x-coordinate, the displacement is the same each second - that is, there is a constant horizontal velocity. Since location 2 has a x-coordinate of 15 m, each consecutive location will be 15 m further than the one before it. The y-coordinate is determined by using the 0.5•a•t2 expression with times of 1 s, 2 s, 3 s, and 4 s. The simplification that g is ~10 m/s2 is used to simplify the math and hilight the concept.
65. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6-meter high building. Determine the horizontal displacement of the projectile.
Answer: x = 27.0 m
This horizontally-launched projectile problem can be (and should be) solved in the same manner as the solution to #60 above. While #60 is broken down for you into nice steps, this problem is not so user-friendly. It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
vix = 12.0 m/s
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = -24.6 m (- means moving down)
viy = 0.0 m/s (its launched horizontally)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time.
Plugging and chugging the above values into this equation yields a time of 2.25 seconds. Now the t value can be combined with the vix and ax value and used in an x-equation
x = vix*t + 0.5*ax*t2
to yield the answer 27.0 m. More examples and discussion of these types of projectile problems are discussed elsewhere.
66. A projectile is launched with an initial speed of 21.8 m/s at an angle of 35.0-degrees above the horizontal.
(a) Determine the time of flight of the projectile.
(b) Determine the peak height of the projectile.
(c) Determine the horizontal displacement of the projectile.
Answer: (a) 2.55 s; (b) 7.98 m; (c) 45.7 m
This non-horizontally-launched projectile problem can be (and should be) solved in the same manner as the solution to #61 above. While #61 is broken down for you into nicely-structured steps, this problem is not so user-friendly. It is strongly recommended that you begin by resolving the initial velocity and angle into initial velocity components using the equations:
vix = vi * cos(theta) viy = vi * sin(theta)
This yields values of vix = 17.9 m/s and viy = 12.5 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
vix = 17.9 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = 0 m (it rises and falls to original height)
viy = 12.5 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find
in which case there will be two solutions: t = 0 s and t = 2.55 s. These two solutions to the equation indicate that the time is 0 s when the vertical displacement (y) is 0 m. This is true before being launched (t = 0 s) and the instant it lands (t=2.55 s). The latter of the two solutions can be used to determine the horizontal displacement (x). Use the equation:
x = vix*t + 0.5*ax*t2
where t is 2.55 s, ax = 0 m/s/s and vix was the first value calculated (using the trigonometric functions). Plugging a chugging the above values into this equation yields the answer of 45.7 m.
Finding the vertical displacement at the peak (ypeak) demands using the original y equation with a time of 1.28 seconds (tup). This time corresponds to the time for one-half of the trajectory - the time at which the projectile will be at its highest or peak position. Substituting the viy, ay and t values into the equation
ypeak = viy*tup + 0.5*ay*tup2
yields a value of 7.98 m for the peak height.
More examples and discussion of these types of projectile problems are discussed elsewhere.
67. A projectile is launched horizontally from the top of a 45.2-meter high cliff and lands a distance of 17.6 meters from the base of the cliff. Determine the magnitude of the launch velocity.
Answer: 9.29 m/s
The best means of starting this problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 17.6 m (the distance horizontally from cliff base)
Vertical Motion
y = -45.2 m (it falls down from the cliff to the ground)
Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there will be two solutions: t = 1.8952 s and t = -1.8952 s. A full parabola which follows the above function would have to locations where the y coordinate is -45.2 m. One would be "forward in time" at 1.8952 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (vix). Use the equation:
x = vix*t + 0.5*ax*t2
where t is 1.8952 s, ax = 0 m/s/s and x = 17.6 m. Plugging and chugging the above values into this equation yields the answer of 9.2865 m/s.
68. Two Glenview students stand on the top of their 3.29-meter second-story deck and launch a water balloon from a home-made winger. The balloon is launched upward at a speed of 45.2 m/s and an angle of 39.1 degrees. The balloon lands in a retention pond whose surface is 2.92 meters below grade. Determine the horizontal distance from launch location to landing location.
Answer: 211 m
This is a non-horizontally-launched projectile problem in which the initial velocity and launch angle are given. Three initial steps are always wisely taken before starting such a problem. First, determine the initial velocity components (vix and viy) using trigonometric functions. Second, construct a diagram of the physical situation. And third, organize known (and unknown) information in an "x-y table." These three steps are taken here. Quickly barging into a solution before giving the problem some pre-analysis often leads to a wasting of much time and ultimately a lot of confusion.
The initial velocity and angle can be resolved into initial velocity components using the
equations:
vix = vi * cos(theta) viy = vi * sin(theta)
This yields values of vix = 35.077 m/s and viy = 28.507 m/s.
A diagram of the physical situation is shown.
Now the known values for each of the variables in the kinematic equations is listed in a table using a column for known horizontal information and a column for known vertical information.
Horizontal Motion
x = ???
vix = 35.077 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = -5.49 m (from initial + to height to final - height)
viy = 28.507 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there will be two solutions: t = -0.1867 s and t = 6.004 s. A full parabola which follows the above function would have two locations where the y coordinate is -5.49 m. One location would be "forward in time" at 6.004 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the horizontal displacement (x).
Now use the equation:
x = vix*t + 0.5*ax*t2
where t = 6.004 s, ax = 0 m/s/s and vix = 35.077 m/s (as originally calculated using the trigonometric functions). Plugging and chugging the above values into this equation yields
the answer of 210.60 m. (Wow! Those boys had better be careful.)
69. A place kicker kicks a football from 39.6 meters from the goal posts. The kick leaves the ground with a speed of 24.8 m/s at an angle of 49.6 degrees. The goal posts are 3.10-meters high.
(a) Determine the amount by which the kick clears the goal posts.
(b) For this given launch velocity, what is the longest field goal (in yards) which could have been kicked? Assume that the football hits the horizontal cross-bar of the posts and bounces through. Given: 1 meter = 3.28 feet.
Answer: (a) 13.7 m; (b) 64.7 yds (measured from kick location to goal posts)
(a) In part a of this problem, the task involves finding the height of the ball (y) when it has traveled a distance of 39.6 meters. The height of the goal posts can be subtracted from this value to determine the amount of clearance.
As is the case in all non-horizontally-launched projectile problems, it should be begun by resolving the initial velocity and angle into initial velocity components using the equations:
vix = vi * cos(theta) viy = vi * sin(theta)
This yields values of vix = 16.073 m/s and viy = 18.886 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 39.6 m (horiz. distance to goal posts)
vix = 16.073 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = ??? (we need to calculate this)
viy = 18.886 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of x-information are now known, an x-equation can be employed to find the time for the football to travel the horizontal distance to the goal posts. One useful equation is
x = vix*t + 0.5*ax*t2
in which case the time is 2.4637 s. The time can now be combined with a y-equations to find the vertical displacement (i.e., height above the ground) when the football has traveled horizontally to the goal posts. Use the equation:
y = viy*t + 0.5*ay*t2
where t = 2.4637 s, ay = -9.8 m/s/s and viy = 18.886 m/s. Plugging a chugging the above values into this equation yields the answer of 16.788 m.
When the ball has traveled a horizontal distance of 39.6 m, it is 16.8 m above the ground. The goal posts are 3.10 m high; so the ball clears the goal posts by 13.7 meters.
(b) Part b of this problem can be done in a similar manner. The task will involve first finding the time for the ball to rise to its peak and then fall back down to a height of 3.10 meters. Then the horizontal displacement can be calculated for this time. The final answer will then need to be converted to yards. The same vix and viy values can be used. Given the new context of the problem, the value for y is now known and x is unknown. The information can be organized in the usual x- and y- table.
Horizontal Motion
x = ???
vix = 16.073 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = 3.10 m (the height of the goal posts)
viy = 18.886 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there will be two solutions: t = 0.1718 s and t = 3.6825 s. The first solution corresponds to the first point along the parabola (during the rise of the football) when the football is at a height of 3.10 m and the second solution is the second point along the parabola (during the fall of the football) when the football is at a height of 3.10 m. The second answer can be used to determine the horizontal displacement (x) of the football. Use the equation:
x = vix*t + 0.5*ax*t2
where t = 3.6825 s, ax = 0 m/s/s and vix = 16.073 m/s. Plugging a chugging the above values into this equation yields the answer of x = 59.189 m. This x value can be converted
to feet by multiplying by the 3.28 ft/m conversion ratio and then converted to yards by dividing by the 3.00 ft/yd conversion ratio. The kicker can kick as 64.7 yard field goal. (In football, it would be referred to as ~47 yard field goal since the goal posts are placed 10 yards behind the goal line and the ball is kicked from about 7 yards behind the line of scrimmage. Field goal distances are are measured from the goal line to the line of scrimmage.)
70. Sammy Sosa clubs a homerun which sails 421 feet and lands on an apartment balcony located a vertical distance of 59 feet above the level of the ball-bat contact location. An observer times the flight to the balcony to take 3.4 seconds.
(a) Determine the velocity (magnitude and angle) at which the ball leaves the bat.
(b) Determine the speed of theball (in miles/hour) when it lands in the bleachers.
Given: 1 m/s = 2.24 mi/hr; 1 meter = 3.28 feet.
Answer: (a) vi = 43.7 m/s at 30.2 degrees; (b) 88.3 mi/hr
(a) For any projectile problem, it always a wise idea to begin the solution with a listing of known and unknown information in an "x-y table." This is shown below.
Horizontal Motion
x = 421 ft = 128.35 m (horiz. distance to balcony)
vix = ???
ax = 0 m/s/s (true for all projectiles)
t = 3.4 s (the ball is a projectile for this long)
Vertical Motion
y = 59 ft = 17.99 m (vert. distance to balcony)
viy = ???
ay = -9.8 m/s/s (true for all projectiles)
t = 3.4 s (the ball is a projectile for this long)
Note that the time of flight is known. Time is a scalar quantity and has no directional component associated with it; one cannot refer to the horizontal time or the vertical time. It is listed in both tables since it can be used with kinematic equations for both the x- and the y-direction.
Since three pieces of x-information are known, an x-equation can be employed to find the initial horizontal velocity. One useful equation is
x = vix*t + 0.5*ax*t2
The initial horizontal velocity (vix) is 37.751 m/s.
There are also three pieces of y-information known. Thus, a y-equation can be used to determine the initial vertical velocity (viy). A good equation is
y = viy*t + 0.5*ay*t2
Plugging a chugging the above values into this equation yields an initial vertical velocity (viy) value of 21.951 m/s.
The ball leaves Sammy Sosa's bat moving upward with a speed of 21.951 m/s and moving horizontally with a speed of 37.751 m/s. These two components of the initial velocity can be used to determine the initial velocity and angle of the baseball after contact with the bat. A diagram is shown at the right. The initial velocity of the ball is represented by the hypotenuse of a right triangle with sides equal to the component values. Thus the Pythagorean theorem can be used to determine the initial velocity of the baseball.
vi 2 = (vix)2 + (viy) 2
vi2 = (37.751 m/s)2 + (21.951 m/s)2
vi 2 = 1906.97 m2/s2
vi = SQRT (1906.97 m2/s2 ) = 43.669 m/s
The angle (theta) of the initial velocity can be determined using a trigonometric function. The tangent function is used here.
(b) In part (a) of this problem, the initial horizontal velocity was determined to be 37.751 m/s. For projectiles, this horizontal velocity does not change during the flight of the projectile. Thus, the projectile strikes the balcony moving with a final horizontal velocity (vfx) of 37.751 m/s. If the final vertical velocity (vfy) can be determined, then it can be used
with the vfx value to determine the final velocity (vf). Several kinematic equations are useable for finding the final vertical velocity (vfy). The following equation will be used:
vfy = viy + ay•t
vfy = 21.951 m/s + (-9.8 m/s/s)•(3.4 s)
vfy = 21.951 m/s - 33.32 m/s
vfy = -11.369 m/s
With the x- and y-components of the final velocity (vf) known, the Pythagorean theorem can be used to determine the final velocity value. A diagram is shown at the right and the calculations are shown below.
vf 2 = (vfx)2 + (vfy) 2
vf2 = (37.751 m/s)2 + (-11.369 m/s)2
vf 2 = 1554.41 m2/s2
vf = SQRT (1554.41 m2/s2) = 39.43 m/s
This value can be converted to miles/hour using the fact that 1 m/s = 2.24 mi/hr. The answer to part b is 88.31 mi/hr.
71. An unfortunate accident occurred on the tollway. A driver accidentally passed through a faulty barricade on a bridge (quite unfortunately). and landed in a pile of hay (quite fortunately). Measurements at the accident scene reveal that the driver plunged a vertical distance of 8.26 meters. The car carried a horizontal distance of 42.1 meters from the location where it left the bridge. If the driver was in a 65 mi/hr speed zone, then determine the amount by which the driver was exceeding the speed limit at the time of the accident. Assume that the contact with the barricade did not slow the car down. (1 m/s = 2.24 mi/hr)
Answer: 72.6 mi/hr
This is an example of a horizontally-launched projectile problem. Like all projectile problems, the best means of starting the problem is to list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = 42.1 m (the horizontal distance which is traveled)
vix = ??? m/s
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = -8.26 m (it falls down from the cliff to the ground)
viy = 0 m/s (it is horizontally launched)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there will be two solutions: t = 1.2983 s and t = -1.2983 s. A full parabola which follows the above function would have two locations where the y coordinate is -8.26 m. One would be "forward in time" at 1.2983 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, the positive answer is the one which we need; it can can be used to determine the initial horizontal velocity (vix). Use the equation:
x = vix*t + 0.5*ax*t2
where t = 1.2983 s, ax = 0 m/s/s and x = 42.1 m. Plugging a chugging the above values into this equation yields the answer of 32.426 m/s. This is the speed at which the car leaves the bridge at the start of its projectile motion. Converting this to mi/hr involves multiplying by the (2.24 mi/hr) / (1 m/s) conversion ratio. The result is 72.6 mi/hr.
(In reality, the car was traveling faster than this speed since the collision with the guard rail likely slowed the car down before it exited the bridge and began its projectile motion.)
72. Cupid wishes to shoot an arrow through the open window of a tall building. The window is 32.8 meters above the ground and Cupid stands 63.6 meters from the base of the building. If Cupid aims the arrow at an angle of 51.5 degrees above the horizontal, with what minimum speed must he fire the arrow in order for it to enter the window?
Answer: 32.7 m/s
Here is an example of a non-horizontally launched projectile problem in which the angle is given but the launch speed is not known. Thus, the x- and y- components of the initial velocity cannot be found. Nonetheless, expressions relating these components to the initial velocity can still be written and used in the problem.
vix = vi * cos(theta) viy = vi * sin(theta)
The usual procedure of listing the known information in a "x-y table" is taken:
Horizontal Motion
x = 63.6 m (horizontal distance to building)
vix = vi * cos(51.5 deg) = 0.6225 • vi
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = 32.8 m (vertical distance from ground to window)
viy = vi * sin(51.5 deg) = 0.7826 • vi
ay = -9.8 m/s/s (true for all projectiles)
As shown in the table, there are only two pieces of x-information and two pieces of y-information given in the problem. Thus, there would seem at first to be insufficient information provided. But as is often the case in a real problem, one can forge ahead using variables in the hopes that there will be a means to introduce another equation which will assist in the solution. So both a horizontal and a vertical displacement equation will be written. (Note that units have been dropped form the solution in order to improve the clarity of the solution.)
Horizontal Displacement
x = vix*t + 0.5*ax*t2
63.6 = (0.6225 • vi ) • t
Vertical Displacement
y = viy*t + 0.5*ay*t2
32.8 = (0.7826 • vi) •t + 0.5 • (-9.8) • t2
Now we have generated two equations with two unknowns and a solution can be found for the initial velocity of the arrow. Equation 1 is used to generate an expression for t in terms of vi. This expression is then substituted into equation 2 in order to solve for the initial velocity (vi). The work is shown below.
From equation 1: t = (63.6) / (0.6225 • vi )
Substituting into equation 2: 32.8 = (0.7826 • vi) •[(63.6) / (0.6225 • vi )] + 0.5 • (-9.8) • [(63.6) / (0.6225 • vi )]2
73. In a Physics demonstration, a projectile is launched from a height of 1.23 m above the ground with a speed of 10.6 m/s at an angle of 30.0 degrees above the horizontal.
(a) What horizontal distance from the launch location will the projectile land?
(b) With what speed does the projectile land?
Answer: (a) x = 11.7 m; (b) vf = 11.7 m/s
(a) As is the case in all non-horizontally-launched projectile problems, it should be begun by resolving the initial velocity (10.6 m/s) and angle (30 degrees) into initial velocity components using the equations:
vix = vi * cos(theta) viy = vi * sin(theta)
This yields values of vix = 9.180 m/s and viy = 5.3 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ??? (the unknown in part a)
vix = 9.180 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
Vertical Motion
y = -1.23 m (vert. distance to floor)
viy = 5.3 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time for the projectile to rise and ultimately fall to the floor. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there are two solutions for the time: t = 1.2780 s and t = -0.1964 s. A full parabola which follows the above function would have two locations where the y coordinate is -1.23 m. One location would be "forward in time" at 1.2780 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, we wish to use the positive time value in our calculations. So t = 1.2780 seconds.
The time can now be combined with a x-equations to find the horizontal displacement (x). Use the equation:
x = vix*t + 0.5*ax*t2
where t = 1.2780 s, ax = 0 m/s/s and vix = 9.180 m/s. Plugging a chugging the above values into this equation yields the answer of 11.739 meters for the horizontal displacement.
(b) The landing speed (vf) of the projectile can be determined from values of the x- and y- component of the final velocity. Since the object being analyzed is a projectile, there is no horizontal acceleration and the final horizontal velocity (vfx) is the same as the initial horizontal velocity (vix) - 9.180 m/s. The final vertical velocity (vfx) can be determined using the following kinematic equation:
vfy = viy + ay•t
vfy = 5.3 m/s + (-9.8 m/s/s)•(1.2780 s)
vfy = -7.2244 m/s
With the x- and y-components of the final velocity (vf) known, the Pythagorean theorem can be used to determine the final velocity value. A diagram is shown at the right and the calculations are shown below.
vf 2 = (vfx)2 + (vfy) 2
vf2 = (9.180 m/s)2 + (-7.2244 m/s)2
vf 2 = 136.462 m2/s2
vf = SQRT (136.462 m2/s2) = 11.682 m/s
74. A car is parked on a cliff overlooking the sea. The cliff is inclined at an angle of 29 degrees below the horizontal. The negligent driver leaves the car in neutral and it begins rolling from rest towards the cliff's edge with an acceleration 4.5 m/s/s. The car moves a linear distance of 57.2 m to the edge of the cliff before plunging into the ocean below. The cliff is 42.2 m above the sea.
(a) Find the speed (in m/s) of the car the moment it leaves the cliff.
(b) Find the time (in seconds) it takes the car to drop to the water below the edge of the cliff.
(c) Find the position (in meters) of the car relative to the base of the cliff when it lands in the sea.
(b) Once the car reaches the edge of the cliff and rolls off, it becomes a projectile with a vertical acceleration of 0 m/s2. The second task involves determining the time of flight of the projectile from the cliff's edge to the water below. Like all non-horizontally launched projectiles, the starting point is to determine the initial horizontal velocity (vix) and the initial vertical velocity (viy). The initial velocity (22.689 m/s) and angle (-29 degrees) can be resolved into initial velocity components using the equations:
vix = vi * cos(theta) viy = vi * sin(theta)
This yields values of vix = 19.844 m/s and viy = -10.999 m/s. Once done, list the known values for each of the variables in the kinematic equations. It is helpful to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
x = ???
Vertical Motion
y = -42.2 m (vert. distance to water)
vix = 19.844 m/s (from trig. function)
ax = 0 m/s/s (true for all projectiles)
viy = -10.999 m/s (from trig. function)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find the time for the projectile to rise and ultimately fall to the floor. One useful equation is
y = viy*t + 0.5*ay*t2
in which case there are two solutions for the time: t = 2.0196 s and t = -4.2644 s. A full parabola which follows the above function would have two locations where the y coordinate is -42.2 m. One location would be "forward in time" at 2.0196 seconds; and the other solution is at a location traced "backwards in time" from the launch time. Of course, we wish to use the positive time value in our calculations. So t = 2.0196 seconds.
(c) The time can now be combined with a x-equations to find the horizontal displacement (x). Use the equation:
x = vix*t + 0.5*ax*t2
where t = 2.0196 s, ax = 0 m/s/s and vix = 19.844 m/s. Plugging a chugging the above values into this equation yields the answer of 40.078 meters for the horizontal displacement.
75. Mike is running to Physics class because he is late at 6.0 m/s. Walking quickly at the same instant from the opposite end of the 20 m long hallway, Derek walks toward Mike (and the physics classroom) at –2.5 m/s. When and where do they meet if they don’t stop at the physics classroom? If the classroom is located 7 m from Derek’s initial location, who makes it there first?
Draw a position-time graph, using the slopes of the lines, you can roughly determine when the two will pass.
dY = dX at catch up points
Since Derek has a negative velocity, the slope of its line on the position time graph must be negative, therefore Derek starts at position 20 m and Mike at 0 m.
dY = vYt + do (Derek’s equation)
dX= vXt (Mike’s equation)
vYt + do = vXt (make equations equal to each other to solve for t since dx = dy)
-2.5t + 20 = 6.0t
t = 2.35 s
Substitute t into one of the equations to find the distance when they pass:
dX = vXt = 6.0*2.35 = 14.1 m from Mike’s starting point.
If the classroom is located 7 m from Derek’s starting point, then it is (20-7 =) 13 m from Mike’s starting point. Therefore Derek will make it to the classroom first as he must pass the classroom to pass Mike.
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