Chapter 2
Motionin One Dimension
©2008 by W.H. Freeman and Company
2-1 Displacement
There is 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.
Distance traveled (dashed line) is measured along the
actual path.
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2-1 Displacement
The displacement is written:
Left:
Right:
Displacement is positive.
Displacement is negative.
©2008 by W.H. Freeman and Company
Average Velocity
Speed: how far an object travels in a given time
interval
Velocity includes directional information:
x2  x1 x
v

t2  t1 t
©2008 by W.H. Freeman and Company
A particle at t1 = -2.0 s is at x1= 3.4 cm and at t2 = 4.5 s is at x2= 8.5 cm.
What is its average velocity? Can you calculate its average speed from
these data?
The average velocity is given by
v 
x
t

8.5cm  3.4 cm
4.5s   2.0s 

5.1cm
6.5s
 0.78cm s
The average speed cannot be calculated. To calculate the average speed, we would
need to know the actual distance traveled, and it is not given.
©2008 by W.H. Freeman and Company
An airplane travels 3100 km at a speed of 790 km/h and then encounters a
tailwind that boosts its speed to 990 km/h for the next 2800 km. What was the
total time for the trip? What was the average speed of the plane for this trip?
The average speed for each segment of the trip is given by
v 
For the first segment,
d
t
t1 
For the second segment,
Thus the total time is
t 
, so
t2 
d1
v1
d2
v2


3100 km
790 km h
2800 km
d
v
for each segment.
 3.924 h
 2.828 h
990 km h
ttot  t1  t2  3.924 h  2.828 h  6.752 h  6.8 h
The average speed of the plane for the entire trip is
v 
d. tot
t tot

3100 km  2800 km
 873.8  8.7  10 km h
6.752 h
©2008 by W.H. Freeman and Company
2
Acceleration
Acceleration is the rate of change of velocity.
v2  v1 v
a

t 2  t1 t
©2008 by W.H. Freeman and Company
Acceleration
Acceleration is a vector, although in onedimensional motion we only need the sign.
The previous image shows positive
acceleration; here is negative acceleration:
©2008 by W.H. Freeman and Company
Acceleration
There is a difference between negative acceleration and
deceleration:
Negative acceleration is acceleration in the negative
direction as defined by the coordinate system.
Deceleration occurs when the acceleration is opposite in
direction to the velocity.
©2008 by W.H. Freeman and Company
Motion with Constant Acceleration
The average velocity of an object during a time
interval t is
The acceleration, assumed constant, is
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2-5 Motion at Constant Acceleration
In addition, as the velocity is increasing at a
constant rate, we know that
Combining these last three equations, we find:
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Motion with Constant Acceleration
We can also combine these equations so as to
eliminate t:
We now have all the equations we need to solve
constant-acceleration problems.
©2008 by W.H. Freeman and Company
At highway speeds, a particular automobile is capable of an acceleration
of about 1.6 m/s2. At this rate, how long does it take to accelerate from 80
km/h to 110 km/h?
The time can be found from the average acceleration,
a 
.
v
t
 1m s 
t 
v
a

110 km h  80 km h
1.6 m s 2

 30 km h  

3.6 km h

1.6 m s 2
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  5.208s  5 s
A world-class sprinter can burst out of the blocks to essentially top speed (of
about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration
of this sprinter, and how long does it take her to reach that speed?
The sprinter starts from rest. The average acceleration is found from
v 2  v02  2a  x  x0  
11.5 m s 2  0
a

 4.408 m
2  x  x0 
2 15.0 m 
v 2  v02
the elapsed time is found by solving
v  v0  at  t 
v  v0
a

11.5 m s  0
4.408 m s
2
 2.61 s
©2008 by W.H. Freeman and Company
s 2  4.41m s 2
A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 s. How
far did it travel in that time?
The words “slowing down uniformly” implies that the car has a constant
acceleration. The distance of travel is found form
x  x0  v t
x  x0  v t
v0  v
2
v0  v
x  x0  v t 
2
but v 
x  x0 
v0  v
2
 21.0 m s  0 m s 
  6.00 sec   63.0 m
2


t 
©2008 by W.H. Freeman and Company
©2008 by W.H. Freeman and Company
Falling Objects
Near the surface of the Earth, all objects
experience approximately the same acceleration
due to gravity.
This is one of the most
common examples of
motion with constant
acceleration.
©2008 by W.H. Freeman and Company
2-7 Falling Objects
In the absence of air
resistance, all objects
fall with the same
acceleration, although
this may be hard to tell
by testing in an
environment where
there is air resistance.
©2008 by W.H. Freeman and Company
©2008 by W.H. Freeman and Company
2-7 Falling Objects
The acceleration due to
gravity at the Earth’s
surface is approximately
9.80 m/s2.
©2008 by W.H. Freeman and Company
A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) How
high does it go? (b) How long is it in the air?
Choose upward to be the positive direction, and take
For the upward path,
v0  22 m s
v0
y0  0 to be at the height where the ball was hit.
at the top of the path, and
a  9.80 m s 2
(a) The displacement can be found from Eq. 2-11c, with x replaced by y .
v 2  v02  2a  y  y0   y  y0 
v 2  v02
2a
 0
0   22 m s 

2 9.80 m s
2
2

 25 m
(b)
The time of flight can be found from Eq. 2-11b, with x replaced by y , using a displacement of 0
for the displacement of the ball returning to the height from which it was hit.
y  y0  v0t  12 at 2  0
 t  v0  12 at   0
 t 0,t 
2v0
a

2  22 m s 
9.80 m s
2
 4.5 s
The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of
t = 4.5 s is the time to return to the original displacement. Thus the answer is t = 4.5 seconds.
©2008 by W.H. Freeman and Company
Integration
Integration is related to the area under
a curve. It is also the inverse of
differentiation
The total area under the curve from t1
to t2 is
t2
lim  v t   v(t ) dt
ti 0
i
i
i
t1
©2008 by W.H. Freeman and Company
Integration
n 1
x
n
x
 dx  n  1 n  1
1
x
 dx  ln x  C
©2008 by W.H. Freeman and Company
Average Velocity
©2008 by W.H. Freeman and Company
Instantaneous Velocity
The instantaneous
velocity is the average
velocity, in the limit as
the time interval
becomes
infinitesimally short.
©2008 by W.H. Freeman and Company
Acceleration
The instantaneous acceleration is the average
acceleration, in the limit as the time interval
becomes infinitesimally short.
©2008 by W.H. Freeman and Company