Motion in a Straight Line

KINEMATICS

- the process of motion is integral to the description of matter characteristics

- all matter is moving - therefore a method must be formulated for accuracy

DISTANCE vs. DISPLACEMENT

1. DISTANCE - defined as the magnitude or length of motion - NO DIRECTION INDICATED symbol = d

2. DISPLACEMENT - magnitude and direction of motion symbol = s

Measurement of Speed

Total distance : the sum of the all changes in position

Time : an interval of change measured in seconds

Position : Separation between the object and the reference point

Types of Speed

Rest : no change in motionInstantaneous: the current speed

of an object at a point of timeAverage : two ways to determine the

“mean” movement Total distance / Total time Sum of the individual speeds / number of speed

measurements

Speed is a scalar quantity

Velocity

Displacement : the change in position based on distance and Direction

Velocity : is the change in Distance per unit time with a specific direction

Velocity is a vector quantityVelocity also has the following conditions:

Average Instantaneous

Constant Speed

An object could be moving at a steady rate.

Thus, its average and instantaneous speed would be the same!

GBS Physics - speed vs. velocity

Question:

A car is traveling at a constant 60mi/hr in a circular path.

Does it have a constant speed?Does it have a constant velocity?

Problem Solving

G – write down what information is given (with units)

U – Identify what you have to find – the unknown (with units)

E – Identify an equation (relationship) that equates the givens and unknown

S – Solve the equation for the unknown (algebraically before substituting any of the "givens" with their units because there will be fewer mistakes in copying the units until the last step).

S – Substitute the givens (with units) and find the answer. 

Sample Problem #1

If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

GIVEN: Ave Speed = 60 km/hr

Distance = 12 kmUNKNOWN: Time - ?

EQUATION : v = d/t

Sample Problem #1

If a car is traveling at an average speed of 60 kilometers per hour, how long does it take to travel 12 kilometers?

SOLVE : v = d/t => t = d/vSUBSTITUTION: t = 12km / 60km/h t = .2 h t = 12 min t = 720 sec

Sample Problem #2

A high speed train travels 454 km in 7200 seconds. What is the train’s average speed?

GIVEN: Time = 7200 s (always a good idea to convert to (base / fundamental units) - time to seconds- if needed!)

Distance = 454 km - (always a good idea to convert to (base / fundamental units) - km to meters- if needed!)

454 km = 454000 m UNKNOWN: Ave Speed - ? EQUATION : v = d/t

Sample Problem #2

A high speed train travels 454 km in 7200 seconds. What is the train’s average speed?

SOLVE : v = d/t => v = d/tSUBSTITUTION: v = 454km / 7200s(with units!) v = 454000m / 7200s v = 63 m/s

Example:If sun light takes about 8 minutes to go from the sun to the earth, how far away from the sun is the earth? Hint: light travels at 186,000 miles per second!!!

GIVEN: Time = 8 min (always a good idea to convert to (base / fundamental units) - time to seconds- if needed!) 8 min = 480 s

Ave Speed = 186,000 mi/s

UNKNOWN: Distance - ?

EQUATION : v = d/t

Example:If sun light takes about 8 minutes to go from the sun to the earth, how far away from the sun is the earth? Hint: light travels at 186,000 miles per second!!!

SOLVE : v = d/t => d = vtSUBSTITUTION: d = 186,000 mi/s X 480s

(with units!) d = 89,000,000mi

Summary

Speed is based on position change relative to the origin

Scalar and Vectors quantities are used to describe motion

Calculations must include formula, substitution of proper units, and final solution in the MKS system.

Relativity of Velocity

Theory, developed in the early 20th century, which originally attempted to account for certain anomalies in the concept of relative motion, but which in its ramifications has developed into one of the most important basic concepts in physical science

Velocity changes when compared to a frame of reference

Acceleration

Acceleration: rate of change of velocity

Acceleration describes how fast an objects speed is changing per

amount of time.

Types of Acceleration

Acceleration , also known as linear acceleration, rate at which the velocity of an object changes per unit of time. A = v/t (Average Acceleration)

Uniform Acceleration : the constant rate of change in Velocity ( Free Fall )9.81 m/s2 (use 10 m/s2 in multiple choice)

If an object has a constant velocity, then its acceleration would be zero.

If an object is slowing down, it is decelerating:

or a NEGATIVE acceleration

Formulas

How to choose the best formula

•Free Fall

•Acceleration due to gravity

•Uniform acceleration

•Distance is not part of the question

•Time is part of the question

How to choose the best formula

•Free Fall•Acceleration due to gravity•Uniform acceleration•Distance is part of the question•Time is part of the question

How to choose the best formula

Choose this formula when the question does not include the TIME

HW p. #2– Q1 - a

A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading:

Wind - 60 mi/hr S

Plane - 100 mi/hr N

Resultant = 40 mi/hr N

HW p. #2– Q1 - b

A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading:

Wind - 60 mi/hr S

Plane - 100 mi/hr S

Resultant = 160 mi/hr S

HW p. #2– Q1 - c

A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading:

Wind - 60 mi/hr S

Plane - 100 mi/hr S

Resultant = 116.6 mi/hr @ S 59º E

a2 + b2 = c2

(60 mi/hr)2 + (100 mi/hr) 2 = c2

C = 116.6 mi/hr = tan -1 (100 mi/hr / 60 mi/hr) = 59º

HW p. #2– Q1 - c

A 60 mi/hr wind is blowing toward the S. What is the resultant velocity of an airplane traveling 100 mi/hr when it is heading:

GIVEN: a. v = 40 mi/hr; b. v = 160 mi/hr; c. v = 116.6 mi/hr ; t= 5 hr

UNKNOWN: d - ? EQUATION : v = d/t SOLVE : v = d/t => d = vt SUBSTITUTION: a. d = (40 mi/hr)(5 hr) = 200 mi (with units!) b. d = (160 mi/hr)(5 hr) = 800 mi c. d = (116.6 mi/hr)(5 hr) = 583 mi

HW p. #2– Q2

Rowboat across a stream flowing @ 3 mi/hr. Boy can row boat @ 4 mi/hr directly across stream.

Water 3 mi/hr

Boat - 4 mi/hr

Resultant = 5 mi/hr @ 36.9º

a2 + b2 = c2

(3 mi/hr)2 + (4 mi/hr) 2 = c2

C = 5 mi/hr = tan -1 (3 mi/hr / 4 mi/hr) = 36.9º

HW p. #3 – Q1

A fish swims at the rate of 2 ft/s. How long will it take this fish to swim 36 ft?

GIVEN: Ave speed = 2 ft/s; d = 36 ft UNKNOWN: time - ? EQUATION : v = d/t SOLVE : v = d/t => t = d/v SUBSTITUTION: t = 36 ft / 2 ft/s (with units!) t = 18 s

HW p. #3 – Q3 part a

A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s?

GIVEN: vi = 0 ft/s; vf = 40 ft/s; t = 10 s

UNKNOWN: a - ? EQUATION : vf = vi + at SOLVE : vf = vi + at => vf - vi = at

vf – vi / t = a SUBSTITUTION: a = vf – vi / t (with units!)

a = (40 ft/s – O ft/s) / 10 s = 4 ft/s2

HW p. #3 – Q3 part b

A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s?

GIVEN: vi = 0 ft/s; vf = 40 ft/s; t = 10 s

UNKNOWN: Ave speed = ?

EQUATION : Ave speed = vf + vi / 2 SOLVE : ave V = vf + vi / 2 SUBSTITUTION: ave V = vf + vi /2ave V = (40 ft/s + 0 ft/s) / 2 = 20 ft/s

HW p. #3 – Q3 part c

A car starts from rest & accelerates up to a velocity of 40 ft/s in 10 s?

GIVEN: vi = 0 ft/s; vf = 40 ft/s; t = 10 s; a = 4 ft/s2; ave v = 20 ft/s

UNKNOWN: d = ?EQUATION : Ave v = d/tSOLVE : Ave v = d/t => d = (ave v)tSUBSTITUTION: d = (ave v)t = (20 ft/s)10 s = 200 ft

Sample Problem #1

A brick falls freely from a high scaffold at a construction site.

1. What is the velocity after 4 seconds?

2. How far does the brick fall in this time?

Solution Given: a = 9.8 m/s2 t = 4s

Vf = 0 m/s + (-9.8 m/s2) ( 4.0 s)

= -39.2 m/s

d = 0 m/s (4s) + .5(-9.8 m/s2) (4s)2

= 0 + .5(-9.8m/s2) (16 s2 )

= -78.4m

What is the velocity What is the velocity after 4 seconds? Find: Vafter 4 seconds? Find: V

How far does the How far does the brick fall in this brick fall in this time? Find: dtime? Find: d

Sample problem #2

An airplane must reach a speed of 71m/s for takeoff. If the runway is 1000m long, what must be the acceleration?

Solution

(71m/s)2 = (0 m/s)2 + 2 (a) ( 1000m)

(-2000m) a = - 5041 m2 / s2

What is the acceleration needed to What is the acceleration needed to take off?take off?

Given: Vi=0 m/s Find: a =?

Vf=71 m/s d=1000m

a = 2.5 m/s 2

Summary

Determine the type of motionList the given informationChoose the best formula from the

Physics formulasSubstitute the proper unitsSolve for the unknown in the equation

GRAPHICAL REPRESENTATION OF VELOCITYslope - the slope of a

displacement vs. time curve would be the velocity

GBS Physics - position vs. time

Constant VelocityPositive Velocity

Positive VelocityChanging Velocity (acceleration)

                                                    

Slow, Rightward (+)

Constant Velocity

Fast, Rightward (+)

Constant Velocity

                          

http://www.physicsclassroom.com/mmedia/kinema/cpv.html

                                                    

Slow, Leftward (-)

Constant Velocity

Fast, Leftward (-)

Constant Velocity

                                                                              

http://www.physicsclassroom.com/mmedia/kinema/cnv.html

GRAPHICAL REPRESENTATION OF ACCELERATION

slope - the slope of a velocity vs. time curve would be the acceleration

GBS Physics - velocity vs time

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/kinema/avd.html

Positive VelocityPositive Acceleration

Positive VelocityZero Acceleration

As previously learned, a plot of velocity-time can be used to determine the acceleration of an object (the slope). We will now learn how a plot of velocity versus time can also be used to determine the displacement of an object.

For velocity versus time graphs, the area bound by the line and the axes represents the displacement.

Determining the Area on a v-t Graph

http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L4e.html

The shaded area is representative of the displacement during from 0 seconds to 6 seconds. This area takes on the shape of a rectangle can be

calculated using the appropriate equation.

Area = b * h Area = (6 s) * (30 m/s)Area = 180 m

The shaded area is representative of the displacement during from 0 seconds to 4 seconds. This area takes on the shape of a triangle can be

calculated using the appropriate equation.

Area = 0.5 * b * h Area = (0.5) * (4 s) * (40 m/s)Area = 80 m

The shaded area is representative of the displacement during from 2 seconds to 5 seconds. This area takes on the shape of a trapezoid can be

calculated using the appropriate equation.

Area = 0.5 * b * (h1 + h2) Area = (0.5) * (3 s) * (20 m/s + 50 m/s)Area = 105 m