MOTION
An object is in motion if its position changes. The mathematical
description of motion is called kinematics. The simplest kind of
motion an object can experience is uniform motion in a straight
line. The object experiences translational motion if it is moving
without rotating.
One-dimensional motion means that objects are only free to
move back and forth along a single line. As a coordinate system
for one-dimensional motion, choose this line to be an x-axis
together with a specified origin and positive and negative
directions.
DISTANCE AND DISPLACEMENT
Distance is the length between any two points in the path of an
object.
Displacement is the length and direction of the change in
position measured from the starting point.
The primary difference between the two is that the distance an
object travels tells you nothing about the direction of travel,
while displacement tells you precisely how far, and in what
direction, from its initial position an object is located.
Distance is the total length of travel and displacement is the net
length of travel accounting for direction.
Distance traveled = AC + CB + BC = 400 m.
Displacement = xf - xi = 200 m
PROBLEM SOLVING GUIDELINES
• Make a list of given quantities (Data)
• Convert units if needed
• Make a sketch if necessary
• Identify what is to be determined
• Always solve the equation for the unknown
• Be consistent with units
• Check that the answer seems reasonable
• Physics problems take practice –
the more you do the easier it will be!
2.1 You leave your home and drive 4.83 km east on Mt Vernon Hwy to go
to the grocery store. After shopping, you go back home by traveling west
on Mt. Vernon Hwy.
a. What distance do you travel during this trip?
xo = 0 km
xgs = 4.83 km
distance = xtotal = xo+ xgs+ xgs
= 0 + 4.83 + 4.83
= 9.66 km
b. What is your displacement?
xo = 0 km
xf = 0 km
displacement = Δx = xf - xo
=0-0
= 0 km
SPEED
If an object takes a time interval t to travel a distance x, then
the average speed of the object is given by:
x
v
t
Units: m/s
2.2 A ship steams at an average speed of 30 km/h.
a. What is the speed in m/s?
v = 30 km/h
30
v
= 8.33 m/s
3.6
b. How far in km does it travel in a day?
v = 30 km/h
t = 24 h
x  vt
x
v
t
= 30(24)
= 720 km
c. How long in hours does it take to travel 500 km?
x = 500 km
x 500
t 
= 16.67 h
v 30
AVERAGE VELOCITY
Average velocity is the displacement divided by the amount of
time it took to undergo that displacement.
The difference between average speed and average velocity is
that average speed relates to the distance traveled while
average velocity relates to the displacement.
2.3 A car travels at 100 km/h for 2 h, at 75 km/h for the next 2 h, and
finally at 80 km/h for 1 h. What is the car’s average velocity for the entire
journey?
v1 = 100 km/h
v2 = 75 km/h
v3 = 80 km/h
t1 = 2 h
t2 = 2 h
t3 = 1 h
Total distance traveled
x=vt
x1 = v1t1 = 100 (2) = 200 km
x2 = 75 (2) = 150 km
x3 = 80 (1) = 80 km
xT = 200 + 150 + 80
= 430 km
Total time
tT = 2 + 2 + 1 = 5 hours
x 430
= 86 km/h

v
5
t
2.4 A runner makes one lap around a 200 m track in a time of 25 s. What
were the runner's
a. average speed and b. average velocity?
x = 200 m
t = 25 s
a.
x 200
v 
= 8 m/s
25
t
b. The run ends at the starting point so
the displacement is zero then
v = 0 m/s
ACCELERATION
Acceleration is the rate of change of velocity. The change in
velocity Δv is the final velocity vf minus the initial velocity vo.
Acceleration happens when:
An object's velocity increases
An object's velocity decreases
An object changes direction
The acceleration of an object is given by:
a
v f  vo
t
Units: m/s2
A positive acceleration means an increase in velocity; a negative
acceleration usually means a decrease in velocity (deceleration)
unless the object is traveling in the negative direction of the
velocity.
EQUATIONS FOR MOTION UNDER CONSTANT
ACCELERATION:
v f  vo  at
vo  vf
x(
)t
2
v f  vo  2ax
2
2
1 2
x  vot  at
2
2.5 An object starts from rest with a constant acceleration of 8 m/s2 along a
straight line.
a. Find the speed at the end of 5 s
v f  vo  at = 0 + 8(5)
vo = 0 m/s
a = 8 m/s2
t=5s
= 40 m/s
b. Find the average speed for the 5 s interval
v
vo  v f
2
0  40

2
= 20 m/s
c. Find the distance traveled in the 5 s
1 2
x  vot  at
2
= 0 + ½(8)(5)2
= 100 m
or:
x  vt
= 20(5)
= 100 m
2.6 A truck's speed increases uniformly from 15 km/h to 60 km/h in 20 s.
a. Determine the average speed
vo = 15/3.6 = 4.17 m/s
vf = 60/3.6 = 16.7 m/s
t = 20 s
v
vo  v f
2
4.17  16.7
= 10.4 m/s

2
b. Determine the acceleration
a
v f  vo
t
16.7  4.17

= 0.63 m/s2
20
c. Determine the distance traveled, all in units of meters and seconds.
x  vt
= 10.4(20)
= 208 m
2.7 A skier starts from rest and slides 9.0 m down a slope in 3.0 s. In what
time after starting will the skier acquire a speed of 24 m/s? Assume that the
acceleration is constant.
vo = 0 m/s, x = 9 m
t = 3 s, vf = 24 m/s
1 2
x  vot  at
2
2 x 2(9)
a  2  2 = 2 m/s2
3
t
t
v f  vo
a
24  0

= 12 s
2
2.8 A bus moving at a speed of 20 m/s begin to slow at a constant rate of 3.0
m/s each second. Find how far it goes before stopping.
vo = 20 m/s
a = - 3 m/s/s
v f  vo  2ax
2
x
2
v v
2
f
2
o
2a
0  202

2( 3)
= 67 m
2.9 A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time
of 5.0 s.
a. Determine the acceleration of the car
vo = 30 m/s
vf = 10 m/s
t=5s
a
v f  vo
t
10  30

5
= - 4 m/s2
b. Determine the distance it moves in the third second
x = vot3 +½at  (vot2 +½at )
2
3
2
2
 vo (t3  t2 )  ½a(t  t )
2
3
= 30(3-2)+ ½(-4)(32-22)
= 20 m
2
2
2.10 The speed of a train is reduced uniformly from 15 m/s to 7.0 m/s while
traveling a distance of 90 m.
a. Calculate the acceleration.
vo = 15 m/s
vf = 7 m/s
x = 90 m
a
v 2f  vo2
2x
v f  vo  2ax
2
2
7 2  152

= - 0.98 m/s2
2(90)
b. How much farther will the train travel before coming to rest, provided
the acceleration remains constant?
x
v 2f  vo2
2a
0  152
= 25 m

2( 0.98)
ACCELERATION DUE TO GRAVITY
All bodies in free fall near the Earth's surface have the same
downward acceleration of:
g = 9.8 m/s2
A body falling from rest in a vacuum thus has a velocity of
9.8 m/s at the end of the first second, 19.6 m/s at the end of the
next second, and so forth. The farther the body falls, the faster
it moves.
A body in free fall has the same downward acceleration
whether it starts from rest or has an initial velocity in some
direction.
Galileo is alleged to have performed freefall experiments by dropping objects off
the Tower of Pisa
Galileo Galilei
(1564-1642)
The presence of air affects the motion of falling bodies partly
through buoyancy and partly through air resistance. Thus two
different objects falling in air from the same height will not, in
general, reach the ground at exactly the same time. Because air
resistance increases with velocity, eventually a falling body
reaches a terminal velocity that depends on its mass, size, shape,
and it cannot fall any faster than that.
FREE FALL
When air resistance can be neglected, a falling body has the
constant acceleration g, and the equations for uniformly
accelerated motion apply. Just substitute a for g.
Sign Convention for
direction of motion:
If the object is thrown
downwards then
g = 9.8 m/s2
If the object is thrown
upwards then
g = - 9.8 m/s2
2.11 A ball is dropped from rest at a height of 50 m above the ground.
a. What is its speed just before it hits the ground?
vo = 0 m/s
y = 50 m
g = 9.8 m/s2
v f  vo2  2 gy  2(9.8)(50) = 31.3 m/s
b. How long does it take to reach the ground?
t
v f  vo
g
313
. 0

= 3.19 s
9.8
2.12 A stone is thrown straight upward and it rises to a height of 20 m.
With what speed was it thrown?
y = 20 m
g = -9.8 m/s2
At highest point vf = 0
vo  v 2f  2 gy  0  2(9.8)(20) = 19.8 m/s
2.13 A stone is thrown straight upward with a speed of 20 m/s. It is caught
on its way down at a point 5.0 m above where it was thrown.
a. How fast was it going when it was caught?
vo = 20 m/s
y=5m
g = -9.8 m/s2
2
v f  vo2  2 gy  20  2( 9.8)(5) = - 17.4 m/s (moving down)
b. How long did the trip take?
t
v f  vo
g
17.4  20

9.8
= 3.8 s
2.14 A baseball is thrown straight upward on the Moon with an initial
speed of 35 m/s. (g = 1.60 m/s2) Find
a. The maximum height reached by the ball
vo = 35 m/s
g = -1.6 m/s2
y
v 2f  vo2
2g
At highest point vf = 0
0  352

2( 16
. )
= 382. 8 m
b. The time taken to reach that height
t
v f  vo
g
0  35
= 21.9 s

1.6
c. Its velocity 30 s after it is thrown
t = 30 s
v f  vo  gt  35  ( 16
. )(30) = - 13 m/s
d. Its velocity when the ball's height is 100 m
y = 100 m
v f  vo2  2 gy  35 2  2(1.6)(100 ) = 30 m/s
2.15 A ball that is thrown vertically upward on the Moon returns to its
starting point in 4.0 s. Find the ball's original speed. (g = 1.60 m/s2)
t=4s
g = -1.6 m/s2
y=0m
vo  ( y 
1
2
y  vo t 
1
1
gt )( ) 
4
t
2
(
1
2
1
2
gt 2
2
( 1.6)( 4 ) )
= 3.2 m/s
2.16 A rock is thrown vertically upward with a velocity of 20 m/s from the
edge of a bridge 42 m above a river. How long does the rock stay in the air?
vo = 20 m/s
y = - 42 m
g = -9.8 m/s2
First, find the velocity of
the rock at the moment
that it hits the river.
2
v f  vo2  2 gy  20  2(9.8)( 42)
= -35 m/s
We select the negative answer because the rock will be moving
toward the river, in the negative direction. Now, use this value
to find t
v f  vo  35  20

= 5.6 s
t
 9 .8
g
GRAPHICAL ANALYSIS OF MOTION
Graphical interpretations for motion along a straight line
(the x-axis) are as follows:
The instantaneous velocity of an object at a certain time is the
slope of the position (displacement) versus time graph at that
time. It can be positive, negative, or zero.
The instantaneous acceleration of an object at a certain time is
the slope of the velocity versus time graph at that time. It can
be positive, negative, or zero.
For constant velocity motion, the x versus t graph is
a straight line.
For constant acceleration motion, the v versus t graph is a
straight line. The area under the curve gives the displacement.
2.17 A car moves in a straight line and its odometer readings are plotted
against time.
a. Find the instantaneous speed of the car at points A and B.
B
x
40

= 0.5 m/s
80
t
Same for A and B
A
b. What is the car's average speed?

v= 0.5 m/s
c. What is its acceleration?
a = 0 m/s2
2.18 An object's one-dimensional motion along the x-axis is graphed in the
figure. Describe its motion.
t = 0-2 s
object is at rest
t = 2-4 s
slope is
30

= 1.5 m/s
42
object moves at constant velocity
t = 4-6 s object is at rest
t = 6-10 s slope is

23
= -1.3 m/s
10  6
object moves in opposite direction
2.19 The vertical motion of an object is graphed in the figure. Describe
its motion qualitatively and find its instantaneous velocity at points A,
B and C.
Object moves fastest at t = 0 as it slows and stops at B, it begins
to fall back downward at increasing speed.
Find its instantaneous velocity at
points A, B and C.
vA
12  3
= 2.3 m/s
40
vB
slope = 0 = 0 m/s
vC
5.5  13
15  8.5
= -1.2 m/s