Post on 23-Nov-2015
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Vector A has a magnitude of 1.35 and a direction angle of 40 above the x axis, and that vector B has a magnitude of 3.05 and a direction angle of 15 below the x axis. Find magnitude and direction of vector C such that
Revision
(a) C = A + B (b) C = A - B
Solution: Practice 1
y
x
|A|=1.35
40O
Solution: Practice 1
y
x
|B|=3.05
35o
325o
(a) C = A + B
Cx = Ax + Bx = 3.53
Cy = Ay + By = -0.88
C = Cx + Cy
C = 3.53 x + (-0.88 y) = 3.53 x 0.88 y
Magnitude
Direction
y
x 14o
C
(a) C = A - B
Cx = Ax - Bx = -1.47 Cy = Ay - By = 2.62 C = Cx + Cy C = -1.47 x + 2.62 y
Magnitude
Direction
y
x 60.7o
29.3o
119.3o
119.3o
C
CHAPTER 2:
One-Dimensional
Kinematics
MAS FIZA MUSTAFA
mfiza@salam.uitm.edu.my
03-3258 4972
013-7133158
Units of chapter:
2.1 Position, Distance and Displacement
2.2 Average speed and velocity
2.3 Instantaneous velocity
2.4 Acceleration
2.5 Motion with constant acceleration
2.6 Applications of The Equations of Motions
2.7 Free falling objects
2.1 Position, Distance and Displacement
To describe the motion of a particle, we need to set up a coordinate system that defines its position.
We make a distinction between distance and displacement.
Displacement (blue line) is how far the object is from its starting point, regardless of how it got there.
Displacement = final position initial position
SI units: meter, m
Distance traveled (dashed line) is measured along the actual path. Distance = total length of travel SI unit: meter, m
2.2 Average speed and velocity
The next step in describing motion is to consider how rapidly an object moves.
The simplest way to characterize the rate of motion is with the average speed
Average speed = distance
elapsed time
(Speed is how far an object travels in a given time interval)
Scalar quantity
Conceptual checkpoint 2.1
You drive 4.00 mi at 30.0 mi/h and then another 4.00 mi at 50.0 mi/h. Is your average speed for the 8.00 mi trip
a) Greater than 40.0 mi/h
b) Equal to 40.0 mi/h
c) Less than 40.0 mi/h
Solution:
Average speed = distance elapsed time We need to calculate the
elapsed time
50.0 mi/h
4.00 mi 8.00 mi
30.0 mi/h
I II
In the 1st 4.00 mi the average speed is 30.0 mi/h, so we can calculate the elapsed time, t1 ( time taken for the car to reach 4.00 mi with the speed of 30.0 mi/h)
50.0 mi/h
4.00 mi 8.00 mi
30.0 mi/h
I II
For the 2nd 4.00 mi the average speed is 50.0 mi/h, so we can calculate the elapsed time,t2 ( time taken for the car to reach another 4.00 mi with the speed of 50.0 mi/h)
Average speed = distance elapsed time
a)Greater than 40.0 mi/h
b)Equal to 40.0 mi/h
c)Less than 40.0 mi/h
Average velocity = displacement elapsed time
(Velocity includes directional information)
Vector quantity
Average velocity tells us how fast something is moving and also the direction of the object.
If an object is moves in the positive direction, then xf > xi then, vav > 0.
On the other hand, if an object moves in the negative direction, then xf < xi then, vav < 0.
Example 2.2
An athlete sprints 50.0m in 8.00s, stops and walk slowly back to the starting line in 40.0s. If the sprint direction is taken to be positive, what are
a) The average sprint velocity?
b) The average walking velocity?
c) The average velocity for the complete round trip?
a)The average sprint velocity?
Sprint ( t = 8.00s )
Average velocity = displacement elapsed time
b) Average walking velocity?
walking ( t = 40.0s )
Average velocity = displacement elapsed time
Negative sign indicates the
motion is to the left
Example 2.3
The position of a runner as a function of time is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runners position changes from x1 = 50.0 m to x2 = 30.5 m. What was the runners average velocity?
-6.5 m/s
The runner moves in the negative-x
direction
How far can a cyclist travel in 2.5 h along a straight road if her average velocity is 18 km/h?
45 KM
The instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short.
Ideally, a speedometer would measure instantaneous velocity; in fact, it measures average velocity, but over a very short time interval.
2.3 Instantaneous velocity
Evaluate the average velocity over shorter and shorter time intervals,
approaching zero in the limit
Note that the instantaneous velocity can be positive, negative or zero just like the average velocity
And just like the average velocity, the instantaneous velocity
is a one-dimensional vector.
The magnitude of the instantaneous velocity is called the instantaneous speed.
On a graph of a particles position vs. time, the instantaneous velocity is the tangent to the curve at any point.
A jet engine moves along an experimental track (which we call the x axis). We will treat the engine as if it were a particle. Its position as a function of time is given by the equation x = At2 + B, where A = 2.10 m/s2 and B = 2.80 m. (a) Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. (b) Determine the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at t = 500 s.
Example 2.4
x = At2 + B A = 2.10 m/s2
B = 2.80 m
t1 = 3.00 s
t2 = 5.00 s
(a) Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s.
x = 2.10t2 + 2.80
(b) Determine the average velocity during this time interval.
Average velocity = displacement elapsed time
(c) Determine the magnitude of the instantaneous velocity at t = 500 s.
Try this!!
The position of a particle as a function of time is given by
x = (-2.00 m/s)t + (3.00 m/s3)t3.
a) Plot x versus t for time from t=0 to t=1.00s.
b) Find the average velocity of the particle from t = 0.150s to t = 0.250s
c) Find the average velocity from t =0.190s to t = 0.210s
a)Plot x versus t for time from t=0 to t=1.00s.
b) Find the average velocity of the particle from t = 0.150s to t = 0.250s
c) Find the average velocity from t =0.190s to t = 0.210s
c) Find the instantaneous velocity at t=0.5s
Answer: 0.25 m/s
2.4 Acceleration
Acceleration is the rate of change of velocity.
A car accelerates along a straight road from rest to 90 km/h in 5.0 s. What is the magnitude of its average acceleration?
Example 2.5
An automobile is moving to the right along a straight highway, which we choose to be the positive x axis. Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m/s, and it takes 5.0 s to slow down to v2 = 5.0 m/s, what was the cars average acceleration?
Example 2.5
If the velocity of a car is non-
zero (v 0), can the
acceleration of the car be zero?
Concept Test: Acceleration
1) yes
2) no
3) depends on the
velocity
Sure it can! An object moving with constant velocity has a non-
zero velocity, but it has zero acceleration because the velocity is
not changing.
When throwing a ball straight up, which of the following is true about its velocity v and its acceleration a at the highest point in its path?
Concept Test: Acceleration
1) both v = 0 and a = 0
2) v 0, but a = 0
3) v = 0, but a 0
4) both v 0 and a 0
5) not really sure
At the top, clearly v = 0 because the
ball has momentarily stopped. But
the velocity of the ball is changing, so
its acceleration is definitely not zero!
Otherwise it would remain at rest!!
2.4 Instantaneous Acceleration
The instantaneous acceleration is the average acceleration in the limit as the time interval becomes infinitesimally short.
A particle is moving in a straight line so that its position is given by the relation x = (2.10 m/s2)t2 + (2.80 m). Calculate (a) its average acceleration during the time interval from t1 = 3.00 s to t2 = 5.00 s, and (b) its instantaneous acceleration as a function of time.
(a)
Example 2.6
(b) its instantaneous acceleration as a function of time.
2.5 Motion with constant acceleration
The average velocity of an object during a time interval t is
The acceleration, assumed constant, is
In addition, as the velocity is increasing at a constant rate, we know that
For constant acceleration
2.6 Applications of the equations of the motions
How long does it take a car to cross a 30.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m/s2?
2.7 Free falling objects
In the absence of air resistance, all objects fall with the same acceleration, although this may be tricky to tell by testing in an environment where there is air resistance.
The acceleration due to gravity at the Earths surface is approximately 9.80 m/s2. At a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.
+y
-y
Since free falling object is about moving in one-direction, we only refer to y-axis
g = -9.81 m/s2 g = 9.81 m/s2
Initial speed in the direction of y-axis
Initial speed in the direction of x-axis
Note that ay = g !!! y = negative (upward) y = positive (downward)
Suppose that a ball is dropped (v0 = 0) from a tower 70.0 m high. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance.
0m
Example 2.7
A ball is thrown straight upward with an initial speed of 25 m/s. Taking upward to be the positive direction, find the speed and the direction of motion of the ball
a) 2.0 seconds
b) 3.0 seconds after it is launched.
Example 2.8
b)
a)
UPWARD
DOWNWARD
Concept Test: Free Falling
You throw a ball straight
up into the air. After it
leaves your hand, at
what point in its flight
does it have the
maximum value of
acceleration?
1) its acceleration is constant
everywhere
2) at the top of its trajectory
3) halfway to the top of its trajectory
4) just after it leaves your hand
5) just before it returns to your hand
on the way down
The ball is in free fall once it is released.
Therefore, it is entirely under the influence of
gravity, and the only acceleration it experiences is
g, which is constant at all points.
Example 2.10
A person steps off the end of a 3.00m high diving board and drops to the water below.
a) How long does it take for the person to reach the water?
b) What is the persons speed on entering the water?
y
Set up coordinate system:
Let positive direction to be
downward
A person steps off the end of a 3.00m high diving board and drops to the water below. a) How long does it take for the person to reach the water? b) What is the persons speed on entering the water?
Summary of Chapter 2
Distance: total length of travel
Displacement: change in position
Average speed: distance / time
Average velocity: displacement / time
Instantaneous velocity: average velocity measured over an infinitesimally small time