Chapter 3, Part 1
Acceleration
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Acceleration
Acceleration (increasing speed) and
deceleration (decreasing speed) should not
be confused with the directions of velocity
and acceleration:
Copyright © 2010 Pearson Education, Inc.
Acceleration
Copyright © 2010 Pearson Education, Inc.
Motion with Constant Acceleration
v f = vi + at
1
vav = (vi + v f )
2
1
s = (vi + v f )t
2
1 2
s = vi t + at
2
2
2
(v f ) = (vi ) + 2as
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v f − vi
t=
a
" v f − vi %
1
s = (vi + v f ) $
'
2
# a &
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 first equation,
v f = vi + at
First, convert 80 .km/h and 110 km/h to m/s.
km 80km
1h
1000m
m
80
=
•
•
= 22.2222
h
1h 3600s 1km
s
km 110km
1h
1000m
m
110
=
•
•
= 30.55556
h
1h
3600s 1km
s
m
m
m
30.55556 = 22.2222 +1.6 2 t
s
s
s
t = 5.2 s
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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 him to
reach that speed?
The sprinter starts from rest. The average acceleration is
found from
2
2
(v f ) = (vi ) + 2as → a =
2
(v f ) 2 − (vi ) 2
2s
11.5m s) − 0
(
=
2 (15.0 m )
2
= 4.408 m s ≈ 4.41m s
2
the elapsed time is found by solving
v f = vi + at → t =
Copyright © 2010 Pearson Education, Inc.
v f − vi
a
11.5m s − 0
=
= 2.61 s
2
4.408 m s
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
! 21.0 m s + 0 m
1
s = (vi + v f )t = #
2
2
"
Copyright © 2010 Pearson Education, Inc.
s$
(
)
& 6.00 sec = 63.0 m
%
Homework
pp. 77-78
pp. 79-82
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Multiple Choice (1-9 odd)
Probs. 5, 9, 19, 31, 37, 55, 63, 67, 73