This document provides an overview of key concepts in physics related to motion, including: - Definitions of terms like speed, velocity, acceleration, displacement, and distance. - Equations for calculating average speed, velocity, acceleration, and distance traveled with constant acceleration. - The relationship between time and velocity/distance for objects experiencing constant acceleration due to gravity. - Centripetal acceleration and the equation relating centripetal acceleration, velocity, and radius of circular motion.

1. Chapter 2

2.  Physics is concerned with the basic principles that
describe how the universe works.
 Physics deals with matter, motion, force, and
energy.
Intro

3.  Classical mechanics
 Waves and sounds
 Thermodynamics
 Electromagnetism
 Quantum mechanics
 Atomic and nuclear physics
 Relativity
Intro

4.  Motion is everywhere – walking, driving, flying,
etc.
 This chapter focuses on definition/discussion of:
speed, velocity, and acceleration.
 There are two basic kinds of motion:
◦ Straight line
◦ Circular
Intro

5.  Position – the location of an object
◦ A reference point must be given in order to define the
position of an object
 Motion – an object is undergoing a continuous
change in position
 Description of Motion – the time rate of change of
position
◦ A combination of length and time describes motion
Section 2.1

6.  In Physical Science ‘speed’ and ‘velocity’ have
different (distinct) meanings.
 Speed is a scalar quantity, with only magnitude
◦ A car going 80 km/h
 Velocity is a vector, and has both magnitude &
direction
◦ A car going 80 km/h north
Section 2.2

7.  Vector quantities may be represented by arrows.
The length of the arrow is proportional to
magnitude.
40 km/h 80 km/h
-40 km/h -80 km/h
Section 2.2
Vectors

8. Note that vectors may be both positive and negative.
Section 2.2

9. ◦ ( where ∆ means ‘change in’)
◦ Over the entire time interval, speed is an average
 Distance – the actual path length traveled
 Instantaneous Speed – the speed of an object at
an instant of time (∆t is very small)
◦ Glance at a speedometer
Section 2.2
distance traveled
time to travel distance
• Average Speed =
• v = d/t or v = ∆d/∆t

10. Section 2.2

11.  Velocity is similar to speed except a direction is
involved.
 Displacement – straight line distance between
the initial and final position w/ direction toward
the final position
 Instantaneous velocity – similar to instantaneous
speed except it has direction
Section 2.2
displacement
total travel time
• Average velocity =

12. Displacement is a vector quantity between two points.
Distance is the actual path traveled.
Section 2.2

13.  Describe the speed of the car above.
 GIVEN: d = 80 m & t = 4.0 s
 SOLVE: d/t = 80 m/4.0 s = 20 m/s = average speed
 Velocity would be described as 20 m/s in the
direction of motion (east?)
Section 2.2
• EQUATION: v = d/t

14.  How far would the car above travel in 10s?
 EQUATION: v = d/t
 REARRANGE EQUATION: vt = d
 SOLVE: (20 m/s) (10 s) = 200 m
Section 2.2

15.  GIVEN:
◦ Speed of light = 3.00 x 108
m/s = (v)
◦ Distance to earth = 1.50 x 108
km = (d)
 EQUATION:
◦ v = d/t  rearrange to solve for t  t = d/v
 t = 0.500 x 103
s = 5.00 x 102
s = 500 seconds
Section 2.2
• SOLVE: t = d/v =
1.50 x 1011
m
3.00 x 108
m/s

16. Section 2.2

17.  GIVEN:
◦ t = 365 days (must convert to hours)
◦ Earth’s radius (r) = 9.30 x 107
miles
 CONVERSION:
◦ t = 365 days x (24h/day) = 8760 h
 MUST FIGURE DISTANCE (d):
 d = ? Recall that a circle’s circumference = 2πr
 d=2πr (and we know π and r!!)
 Therefore  d = 2πr = 2 (3.14) (9.30 x 107
mi)
Section 2.2

18.  SOLVE EQUATION:
 ADJUST DECIMAL POINT:
Section 2.2
• d/t = 2 (3.14) (9.30 x 107
mi) = 0.00667 x 107
mi/h
8760 h
• v = d/t
• v = avg. velocity = 6.67 x 104
mi/h = 66,700 mi/h

19.  Radius of the Earth = RE = 4000 mi
 Time = 24 h
 Diameter of Earth = 2πr = 2 (3.14) (4000mi)
◦ Diameter of Earth = 25,120 mi
 v = d/t = (25,120 mi)/24h = 1047 mi/h
Section 2.2

20.  Changes in velocity occur in three ways:
◦ Increase in magnitude (speed up)
◦ Decrease in magnitude (slow down)
◦ Change direction of velocity vector (turn)
 When any of these changes occur, the object is
accelerating.
 Faster the change  Greater the acceleration
 Acceleration – the time rate of change of velocity
Section 2.3

21.  A measure of the change in velocity during a
given time period
Section 2.3
• Avg. acceleration =
change in velocity
time for change to occur
• Units of acceleration = (m/s)/s = m/s2
• In this course we will limit ourselves to situations
with constant acceleration.
• a = =
∆v
t
vf – vo
t
(vf = final & vo = original)

22.  As the velocity
increases, the
distance traveled
by the falling object
increases each
second.
Section 2.3

23.  A race car starting from rest accelerates uniformly
along a straight track, reaching a speed of 90
km/h in 7.0 s. Find acceleration.
 GIVEN: vo = 0, vf = 90 km/h, t = 7.0s
 WANTED: a in m/s2
Section 2.3
• vf = 90 km/h = 25 m/s)( /
/278.0
hkm
sm
• a = = = 3.57 m/s2
vf –vo
t
25 m/s – 0
7.0 s

24.  Rearrange this equation:
 at = vf – vo
 vf = vo + at (solved for final velocity)
 This equation is very useful in computing final
velocity.
Section 2.3
vf – vo
t
• Remember that a =

25.  If the car in the preceding example continues to
accelerate at the same rate for three more
seconds, what will be the magnitude of its velocity
in m/s at the end of this time (vf )?
 a = 3.57 m/s2
(found in preceding example)
 t = 10 s
 vo = 0 (started from rest)
 Use equation: vf = vo + at
 vf = 0 + (3.57 m/s2
) (10 s) = 35.7 m/s
Section 2.3

26. Section 2.3
Acceleration (+) Deceleration (-)

27.  Special case
associated with
falling objects
 Vector towards the
center of the earth
 Denoted by “g”
 g = 9.80 m/s2
Section 2.3

28. 0
10
20
30
40
50
60
70
0 2 4 6 8
Time in seconds
Velocity
The increase in velocity is linear.
Section 2.3

29.  If frictional effects (air resistance) are
neglected, every freely falling object
on earth accelerates at the same rate,
regardless of mass. Galileo is credited
with this idea/experiment.
 Astronaut David Scott
demonstrated the principle on the
moon, simultaneously dropping a
feather and a hammer. Each fell at
the same acceleration, due to no
atmosphere & no air resistance.
Section 2.3

30.  Galileo is also
credited with using
the Leaning Tower
of Pisa as an
experiment site.
Section 2.3

31.  d = ½ gt2
 This equation will compute the distance (d) an
object drops due to gravity (neglecting air
resistance) in a given time (t).
Section 2.3

32.  A ball is dropped from a tall building. How far
does the ball drop in 1.50 s?
 GIVEN: g = 9.80 m/s2
, t = 1.5 s
 SOLVE:
d = ½ gt2
= ½ (9.80 m/s2
) (1.5 s)2
= ½ (9.80 m/s2
) (2.25 s2
) = ?? m
Section 2.3

33.  A ball is dropped from a tall building. How far
does the ball drop in 1.50 s?
 GIVEN: g = 9.80 m/s2
, t = 1.5 s
 SOLVE:
d = ½ gt2
= ½ (9.80 m/s2
) (1.5 s)2
= ½ (9.80 m/s2
) (2.25 s2
) =
Section 2.3
11.0 m

34.  What is the speed of the ball in the previous
example 1.50 s after it is dropped?
 GIVEN: g = a = 9.80 m/s2
, t = 1.5 s
 SOLVE:
 vf = vo + at = 0 + (9.80 m/s2
)(1.5 s)
 Speed of ball after 1.5 seconds = 14.7 m/s
Section 2.3

35.  Distance is
proportional to t2
– Remember that (d
= ½ gt2
)
 Velocity is
proportional to t
– Remember that (vf
= at )
Section 2.3

36. 0
50
100
150
200
0 2 4 6 8
Velocity
Distance
Time in seconds
Section 2.3

37.  Acceleration due to
gravity occurs in
BOTH directions.
◦ Going up (-)
◦ Coming down (+)
 The ball returns to
its starting point
with the same
speed it had
initially. vo = vf
Section 2.3

38.  Although an object in uniform circular motion has
a constant speed, it is constantly changing
directions and therefore its velocity is constantly
changing directions.
◦ Since there is a change in direction there is a change in
acceleration.
 What is the direction of this acceleration?
 It is at right angles to the velocity, and generally
points toward the center of the circle.
Section 2.4

39.  Supplied by friction of the tires of a car
 The car remains in a circular path as long as there
is enough centripetal acceleration.
Section 2.4

40.  This equation holds true for an object moving in a
circle with radius (r) and constant speed (v).
 From the equation we see that centripetal
acceleration increases as the square of the speed.
 We also can see that as the radius decreases, the
centripetal acceleration increases.
Section 2.4
v2
r
• ac =

41.  Determine the magnitude of the centripetal
acceleration of a car going 12 m/s on a circular
track with a radius of 50 m.
Section 2.4

42.  Determine the magnitude of the centripetal
acceleration of a car going 12 m/s on a circular
track with a radius of 50 m.
 GIVEN: v = 12 m/s, r = 50 m
 SOLVE:
Section 2.4
• ac = = = 2.9 m/s2
v2
r
(12 m/s)2
50m

43.  Compute the centripetal acceleration in m/s2
of the
Earth in its nearly circular orbit about the Sun.
 GIVEN: r = 1.5 x 1011
m, v = 3.0 x 104
m/s
 SOLVE:
Section 2.4
• ac = = =
v2
r
(3.0 x 104
m/s)2
1.5 x 1011
m
9.0 x 108
m2
/s2
1.5 x 1011
m
• ac = 6 x 10-3
m/s2

44.  An object thrown horizontally combines both
straight-line and vertical motion each of which act
independently.
 Neglecting air resistance, a horizontally projected
object travels in a horizontal direction with a
constant velocity while falling vertically due to
gravity.
Section 2.5

45.  An object thrown
horizontally will fall at
the same rate as an
object that is dropped.
Multiflash photograph of two balls
Section 2.5
Copyright © Standard HMCO copyright line

46. Section 2.5

47. Combined Horz/Vert.
Components
Vertical
Component
Horizontal
Component
+=
Section 2.5

48. Section 2.5

49. Section 2.5

50. Section 2.5

51.  Angle of release
 Spin on the ball
 Size/shape of ball/projectile
 Speed of wind
 Direction of wind
 Weather conditions (humidity, rain, snow, etc)
 Field altitude (how much air is present)
 Initial horizontal velocity (in order to make it out of
the park before gravity brings it down, a baseball
must leave the bat at a high velocity)
Section 2.5

52.  d = ½at2
(distance traveled, starting from rest)
 d = ½gt2
(distance traveled, dropped object)
 g = 9.80 m/s2
= 32 ft/s2
(acceleration, gravity)
 vf = vo + at (final velocity with constant a)
 ac = v2
/r (centripetal acceleration)
Review
∆v
t
vf –vo
t
• a = = (constant acceleration)
• v = d/t (average speed)