# phy ch 2 - wbm ```Motion Along a Straight
Line
Physics chapter 2
Physics chapter 2
1
Definitions



Mechanics – the study of the
relationships among force, matter and
motion
Kinematics – the description of motion
Dynamics – the relation of motion to its
causes
Physics chapter 2
2
Average velocity


Vector quantity
Change in x (displacement) divided by
change in time
x2  x1 x
vav 

t2  t1 t


 means “change in”
Always take final minus initial
Physics chapter 2
3
Average velocity



Only depends on total displacement
Not the details of the motion
On a x-t graph (position as a function of
time), it is the slope of the line connecting
the starting point and the ending point.
Physics chapter 2
4
Sign of velocity




The sign (positive or negative) of velocity
indicates direction
It depends on how you set up your
coordinate system
You can choose your coordinate system,
but you must be consistent.
Since t is always positive, v has the
same sign as x
Physics chapter 2
5
Velocity units





Must be displacement (length) units
divided by time units
m/s
km/h
mi/h
ft/s
Physics chapter 2
6
Example


Jeff leaves his house at 10 am and travels
240 km east. He stops for lunch at noon. At
12:30 he gets back in his car and travels
another 100 km east. He arrives at his
destination at 1:30 pm.
What was his average velocity for the first part
of his trip, the second part of his trip, and the
whole trip?
Physics chapter 2
7
Example

First part of trip
t1  10 : 00
t2  12 : 00
240 km
x

vav 
2.0 h
t
x  240 km
km
 120
h
Physics chapter 2
8
Example

Second part of trip
t1  12 : 30
vav. x
t2  1 : 30
x 100 km


1h
t
x  100 km
km
 100
h
Physics chapter 2
9
Example

Whole trip
t1  10 : 00
vav , x
t2  1 : 30
x 340 km


3.5 h
t
x  340 km
km
 97
h
Physics chapter 2
10
Instantaneous velocity


Can tell us the details of a particle’s
motion at a given time.
In physics, an instant has no duration – a
single point in time
Physics chapter 2
11
Instantaneous velocity


Take the limit of average velocity as t
goes to zero
The derivative of x with respect to t
x dx
v  lim

t 0 t
dt
Physics chapter 2
12
Instantaneous velocity


On an x-t graph, it is the slope of the
tangent line to the curve at a given point
In other words, the derivative
Physics chapter 2
13
vocab clarification


“velocity” refers to instantaneous velocity
unless stated otherwise
“speed” is distance divided by time – a
scalar – has nothing to do with direction
Physics chapter 2
14
Example

A car travels along a straight road so
that its distance from the starting point is
given by
m 2 
m 3

xt   1.50 2 t   0.050 3 t
s 
s 



What is its instantaneous velocity at
t = 3 s?
Physics chapter 2
15
Example
m
m


2
3
x3 s   1.50 2 3 s    0.050 3 3 s 
s 
s 


x3 s  12.15 m
t  0.1 s
x3.1 s  12.92545m
12.92545 - 12.15
v
0.1
v  7.75 m
s
t  0.01 s
x3.01s  12.226605m
12.226605 - 12.15
v
0.01
v  7.66 m
s
Physics chapter 2
16
Example
t  0.001 s
x3.001s  12.15765m
12.15765 - 12.15
v
0.001
v  7.65 m
s
t  0.0001 s
x3.0001s  12.150765m
12.150765 - 12.15
v
0.0001
v  7.65 m
s
Physics chapter 2
17
Shortcut with calculus
m 2 
m 3

xt   1.50 2 t   0.050 3 t
s 
s 


m
m 2


dx
 21.50 2 t  3 0.050 3 t
v
s 
s 
dt


m
m


2
 21.50 2 3 s   3 0.050 3 3 s 
s 
s 


m
 7.65
s
Physics chapter 2
18
Average acceleration


A vector quantity
The rate of change of velocity with time
v2  v1 v
aav 

t2  t1 t
Physics chapter 2
19
Instantaneous acceleration

The derivative of v with respect to t
v dv
a  lim

t 0 t
dt
Physics chapter 2
20
Acceleration on a v-t graph

Average acceleration is the slope of a
line connecting a beginning and ending
point.

Instantaneous acceleration is the slope
of a tangent line at a given point.
Physics chapter 2
21
Acceleration on a x-t graph

If the graph is concave up,

If the graph is concave down,

If the graph has no curvature (an
inflection point),
• Then the acceleration is positive.
• Then the acceleration is negative.
• Then the acceleration is zero.
Physics chapter 2
22
Sign of acceleration


The sign of the acceleration doesn’t tell
us whether the object is speeding up or
slowing down.
We also need to know the sign of the
velocity
Physics chapter 2
23
Sign of acceleration




If v and a have the same sign,
Then the object is speeding up.
If v and a have opposite signs,
Then the object is slowing down.
Physics chapter 2
24
Position and acceleration

Acceleration is the second derivative of
position
dv
a
dt
dx
v
dt
d  dx  d x
a   2
dt  dt  dt
2
Physics chapter 2
25
Very important!
Motion with constant
acceleration
v  v0  at
1 2
x  x0  v0t  at
2
v  v  2ax  x0 
2
2
0
 v0  v 
x  x0  
t
 2 
Physics chapter 2
26
CAUTION

These equations only work for constant
acceleration along a straight line!
Physics chapter 2
27
Example

On a highway at night you see a stalled
vehicle and brake your car to a stop with
an acceleration of magnitude 5 m/s2. If
67 mph), what is your stopping distance?
Physics chapter 2
28
Example
m
a  5 2
s
m
v0  30
s
m
v0
s
x0  0 m
v  v  2ax  x0 
2
2
0
0  v  2ax
2
0
Physics chapter 2
29
Example
 2ax  v
2
0
2
0
v
x
 2a
2
 m
 30 
s

x
m

 2  5 2 
s 

Physics chapter 2
30
Example
2
m
900 2
s
x
m
10 2
s

x  90 m
Physics chapter 2
31
Example



A motorist passes a school-crossing corner,
where the speed limit is 10 m/s, traveling with
constant velocity of 20 m/s. A police officer on
a motorcycle, stopped at the corner, starts off
in pursuit with constant acceleration of 2.0 m/s.
How much time elapses before the motorcycle
overtakes the car?
How fast is the motorcycle traveling at this
time?
Physics chapter 2
32
Free fall

Galileo proposed that all objects fall with
the same acceleration, regardless of
their weight.
Physics chapter 2
33
Acceleration due to gravity



When the effects of air can be neglected,
Galileo is right.
The magnitude of the acceleration due to
gravity is denoted with the letter g.
Near the surface of the earth,
g = 9.8 m/s2.
Physics chapter 2
34
Example


While standing in an elevator, you see a
screw fall from the ceiling. The ceiling is
3 m above the floor. The elevator is not
moving.
How long does it take for the screw to hit
the floor?
Physics chapter 2
35
Example
1 2
x  x0  v0t  at
2
x0

x0
v0  0
a  g
1 2
0  x0  gt
2
Physics chapter 2
36
Example
1 2
gt  x0
2
2 x0
t 
g
23 m
t 
m
9.8 2
s
t  0.612s
2
2
2
2
t  0.78 s
Physics chapter 2
37
Example

What if the elevator is moving upward
with a constant acceleration of
magnitude 4.0 m/s2?
Physics chapter 2
38
Example
1
2
x f  x0 f  v0 f t  a f t
2
1 2
xs  x0 s  v0 s t  as t
2
x0 f  0
xs
as   g

v0s  v0 f  v0
xf
x0
Physics chapter 2
39
Example
1
2
x f  v0t  a f t
2
1 2
xs  x0 s  v0t  gt
2
at timet
x f  xs
1
1 2
2
v0t  a f t  x0 s  v0t  gt
2
2
Physics chapter 2
40
Example
1
1 2
2
a f t  x0 s  gt
2
2
1
1 2
2
a f t  gt  x0 s
2
2
1
a f  g t 2  x0 s
2
Physics chapter 2
41
Example
2 x0 s
t 
af  g
2
1
m
a f  4.0 2
s
m
g  9 .8 2
s
23.0 m
2
t1 
m
m
4.0 2  9.8 2
s
s
Physics chapter 2
x0s  3.0 m
42
Example
6.0 m
t 
m
13.8 2
s
2
t  0.435s
2
2
t  0.66 s

Does it make sense that the time is shorter?
Why or why not?
Physics chapter 2
43
Relative velocity


No homework problems, just a short
discussion.
When you are moving, the velocity of
something looks different than when you
are standing still.
Physics chapter 2
44
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