PHYSICAL SCIENCE PSCI 1
CHAPTER 2: MOTION
2
Physics:
The Most Fundamental Physical
Science
Physics is concerned with the basic principles that
describe how nature works.
Physics deals with matter, motion, force, and
energy.
Intro
3
Physics – Areas of Study
Classical mechanics Waves and sounds
Thermodynamics
Electromagnetism
Optics
Relativity
Quantum mechanics
Atomic and nuclear physics
Intro
4
Motion
This chapter focuses on definition/discussion of:
speed, velocity, and acceleration.
There are two basic kinds of motion:
Straight line
Circular (more generally, curvilinear)
Intro
5
Defining Motion
Position – the location of an object
A reference point and a measurement scale 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
Speed and Velocity
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
Vectors
Vector quantities may be represented by arrows.
The length of the arrow is proportional to
magnitude.
40 km/h
80 km/h
Section 2.2
8
Vectors
Note that for motion along one axis only, vectors may be
expressed as positive or negative numbers to represent
direction.
Section 2.2
9
Speed
Average Speed
Δd d − d0
v=
=
Δt
t − t0
d
v=
t
Where:
d = distance, t = time, v = speed
Greek ∆ means “change in”
∆d means change in distance
∆t means change in time
Distance: the actual path length traveled
Instantaneous Speed: v, the speed of an object at
an instant of time (Δt is very small)
Glance at a speedometer
Section 2.2
10
Instantaneous speed
Section 2.2
11
Velocity
Displacement – straight line distance between the initial
and final position with direction toward the final position
!
Instantaneous velocity – similar to instantaneous speed
except it has direction
Velocity is similar to speed except a direction is
involved.
Section 2.2
12
Displacement and Distance
Displacement is a vector quantity between two points.
Distance is the actual path traveled.
Section 2.2
13
Constant Velocity - Example
Describe the speed of the car above.
GIVEN: d = 80 m & t = 4.0 s
d 80 m
m
v= =
= 40
= 40 m/s
t 4.0 s
s
Section 2.2
14
Constant Velocity – Confidence Exercise
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
Example: How long does it take sunlight to reach Earth?
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
d
1.5 × 1011 m
t= =
8
v 3.00 × 10 m/s
3
2
t = 0.5 × 10 s = 5.00 × 10 s = 500s
Section 2.2
16
Sunlight
Section 2.2
17
What is the average speed in mi/h of the Earth revolving around the Sun?
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 CALCULATE DISTANCE (d):
Recall that a circle’s circumference = 2πr
d = 2πr (and we know π and r )
7
Therefore: d = 2πr = 2 (3.14) (9.30 x 10 mi)
Section 2.2
18
What is the average speed in mi/h of the Earth revolving around the Sun?
SOLVE EQUATION:
d
v=
t
7
2 ( 3.14 ) ( 9.3 × 10 mi)
4
v=
= 6.67 × 10 mi/h
8, 760 h
Or: v = 66,700 mi/h
Section 2.2
19
Confidence Exercise: What is the average speed in mi/h of a person at the equator as a result of the Earth’s rotation?
Radius of the Earth = 4000 mi
Time = 24 h (actually, 23 h, 56 min, 4 s)
Diameter of Earth = 2πr = 2 (3.14) (4000mi)
Diameter of Earth = 25,120 mi
v = d/t = (25,120 mi)/24h = 1,047 mi/h
Section 2.2
20
Acceleration
Changes in velocity occur in three ways:
Change in speed
Change in direction
When either or both 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
Acceleration
A measure of the change in velocity during a given
time period
Δv v − v0
a=
=
Δt t − t 0
The units of acceleration are:
velocity m/s
2
=
= m/s
time
s
Section 2.3
22
2
Constant Acceleration of 9.8 m/s
As the velocity
increases, the distance
traveled by the falling
object increases each
second.
Section 2.3
23
Finding Acceleration - Example
A race car starting from rest accelerates uniformly
along a straight track, reaching a speed of 90 km/h in 7.0 s. Find its average acceleration.
GIVEN: v0 = 0, vf = 90 km/h, t0 = 0, tf = 7.0s
WANTED: average acceleration, a, in m/s2
Δv v f − v0 90 km/h
a=
=
=
Δt t f − t 0
7.0 s
90 km ⎛ 10 m ⎞ ⎛ 1 h
a=
⎜⎝
⎜
⎟
7.0 s h ⎝ 1 km ⎠ 3600
3
⎞
2
⎟⎠ = 3.6 m/s
s
Section 2.3
24
Using the equation for acceleration
v f − v0
a=
Δt
a ( Δt ) = v f − v0
v f = a ( Δt ) + v0
This last equation is very useful in computing final
velocity.
v f = at + v0
You could also write
if it’s clear to you what t represents.
Section 2.3
25
Finding Acceleration – Confidence Exercise
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 v0 = 0 (started from rest)
Use equation: vf = v0 + at
vf = 0 + (3.57 m/s2) (10 s) = 35.7 m/s
Section 2.3
26
Acceleration is a vector quantity, since velocity is a vector quantity
Acceleration (+)
Deceleration (-)
Section 2.3
27
Constant Acceleration = Gravity = 9.8 m/s2
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
Graph – Acceleration due to Gravity
Acceleration due to Gravity
Velocity, m/s
40
30
20
10
0
1
2
3
4
Time, s
The increase in velocity is linear because the acceleration is constant.
Section 2.3
29
Frictional (Dissipative) Effects
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
Figure 2
Galileo is also credited with
using the Leaning Tower of
Pisa as an experiment site.
Section 2.3
31
What about the distance a dropped object will travel in a given time?
Recall:
v f = at + v0
For v0 = 0, a = g:
v f = gt
Average velocity:
Distance fallen
in time t.
v
(
v=
f
+ v0
2
)=v
1
= gt
2 2
f
d = vt
1
1 2
d = gt × t = gt
2
2
Section 2.3
32
What about the distance a dropped object will travel in a given time?
1 2
d = gt
2
This equation will compute the distance (d) an object
drops due to gravity (neglecting air resistance) in a
given time (t).
Units check:
⎛ m⎞ 2
m= ⎜ 2⎟ s
⎝s ⎠
( )
33
Solving for distance dropped - Example
An object is dropped from a tall building. How far
does the object 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
34
Solving for distance dropped - Example
An object is dropped from a tall building. How far
does the object 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) = 11.0 m
Section 2.3
35
Solving for final speed – Confidence Exercise
What is the speed of the object in the previous
example 1.50 s after it is dropped?
GIVEN: g = a = 9.80 m/s2, t = 1.5 s
SOLVE:
vf = v0 + at = 0 + (9.80 m/s2)(1.5 s)
Object’s speed after 1.5 seconds = 14.7 m/s
Section 2.3
36
Constant Acceleration = Gravity 9.8 m/s2
Distance is
proportional to t2 Remember that (d = ½ gt2)
Velocity is proportional
to t Remember that (vf = gt )
Section 2.3
37
v proportional to time (t) 2
d proportional to time squared (t )
velocity
distance
Velocity, m/s; Distance, m
80
60
40
20
0
0
1
2
Time, s
38
3
4
Up and Down – Gravity slows the ball, then speeds it up
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. v0 = vf
Section 2.3
39
Acceleration In Uniform Circular Motion
Although an object in uniform circular motion has
a constant speed, it is constantly changing
directions and therefore its velocity is constantly
changing. Since there is a change in direction there is an acceleration.
What is the direction of this acceleration?
It is at right angles to the velocity, and points
toward the center of the circle.
Section 2.4
40
Centripetal (“center-seeking”)
Acceleration
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
41
Centripetal Acceleration
2
v
ac =
r
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
42
Finding Centripetal Acceleration Example
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:
12 m/s )
v
(
2
ac = =
= 2.9 m/s
r
50 m
2
2
Section 2.4
43
Finding Centripetal Acceleration Confidence Exercise
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:
ac
3.0 × 10
(
=
4
m/s
)
2
1.5 × 10 m
−3
2
ac = 6.0 × 10 m/s
11
9.0 × 10 8 m 2 /s 2
=
11
1.5 × 10
Section 2.4
44
Projectile Motion
An object thrown horizontally combines both straight-line
and vertical motion each of which act independently.
(Think of space as providing 3 independent degrees
of freedom.)
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
Figure 2.14(b)
An object thrown
horizontally will fall at the
same rate as an object
that is dropped.
Multi-flash photograph of two objects
Section 2.5
46
The velocity in the horizontal direction does not affect the velocity and acceleration in the vertical direction.
Section 2.5
47
Projected at an angle (not horizontal)
Combined Horz./Vert.
=
Components
Vertical
+
Component
Horizontal
Component
Section 2.5
48
In throwing a football the horizontal velocity remains
constant but the vertical velocity changes like that
of an object thrown upward.
Section 2.5
49
If air resistance is neglected, projectiles have symmetric paths and the maximum range is attained at 45 degrees.
Section 2.5
50
Under real-world conditions, air resistance
causes the paths to be non-symmetric. Air
resistance reduces the horizontal velocity.
Section 2.5
51
Projectiles – Athletic considerations
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
Section 2.5
52
Derived
Definitions
Important Equations – Chapter 2
Δd d − d0
v=
=
Δt
t − t0
Average speed:
d
v=
t
Acceleration:
Δv v − v0
a=
=
Δt t − t 0
Final velocity:
v f = at + v0
Distance fallen:
1 2
d = gt
2
Centripetal
acceleration:
a = g fof the case of
gravity
g = 9.80 m/s2 = 32 ft/s2
(acceleration, gravity)
2
v
ac =
r
Review
53