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Chapter 3Acceleration and Free Fall
the acceleration between t=1.0 and 1.5 s.
Answer: Reading from the graph, it looks like the velocity is
about -1 m/s at t=1.0 s, and around -6 m/s at t=1.5 s. The
acceleration, figured between these two points, is
a=
.
v
.
t
=(–6m/s)–(–1m/s)
(1.5s)–(1.0s)
=–10m/s
2
.
Even though the ball is speeding up, it has a negative
acceleration.
Another way of convincing you that this way of handling the plus and
minus signs makes sense is to think of a device that measures acceleration.
After all, physics is supposed to use operational definitions, ones that relate
to the results you get with actual measuring devices. Consider an air
freshener hanging from the rear-view mirror of your car. When you speed
up, the air freshener swings backward. Suppose we define this as a positive
reading. When you slow down, the air freshener swings forward, so we’ll
call this a negative reading on our accelerometer. But what if you put the car
in reverse and start speeding up backwards. Even though you’re speeding
up, the accelerometer responds in the same way as it did when you were
going forward and slowing down. There are four possible cases:
motion of car
accelerom-
eter
swings
slope of
v-t
graph
direction of
force acting
on car
forward, speeding upbackward+forward
forward, slowing downforward-backward
backward, speeding upforward-backward
backward, slowing downbackward+forward
Note the consistency of the three right-hand columns — nature is
trying to tell us that this is the right system of classification, not the left-
hand column.