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