78
v
(m/s)
t (s)
0123
0
10
20
30
2
4
0
1
2
3
78910
t (s)
Chapter 3Acceleration and Free Fall
3.2Acceleration
Definition of acceleration for linear v-t graphs
Galileo’s experiment with dropping heavy and light objects from a
tower showed that all falling objects have the same motion, and his in-
clined-plane experiments showed that the motion was described by v=ax+v
o
.
The initial velocity v
o
depends on whether you drop the object from rest or
throw it down, but even if you throw it down, you cannot change the slope,
a, of the v-t graph.
Since these experiments show that all falling objects have linear v-t
graphs with the same slope, the slope of such a graph is apparently an
important and useful quantity. We use the word acceleration, and the
symbol a, for the slope of such a graph. In symbols, a=
.
v/
.
t. The accelera-
tion can be interpreted as the amount of speed gained in every second, and
it has units of velocity divided by time, i.e. “meters per second per second,”
or m/s/s. Continuing to treat units as if they were algebra symbols, we
2
.” Acceleration can be a useful quantity for
describing other types of motion besides falling, and the word and the
symbol “a” can be used in a more general context. We reserve the more
specialized symbol “g” for the acceleration of falling objects, which on the
surface of our planet equals 9.8 m/s
2
. Often when doing approximate
calculations or merely illustrative numerical examples it is good enough to
use g=10 m/s
2
, which is off by only 2%.
Example
Question: A despondent physics student jumps off a bridge, and
falls for three seconds before hitting the water. How fast is he
going when he hits the water.
Solution: Approximating g as 10 m/s
2
, he will gain 10 m/s of
speed each second. After one second, his velocity is 10 m/s,
after two seconds it is 20 m/s, and on impact, after falling for
three seconds, he is moving at 30 m/s.
Example: extracting acceleration from a graph
Question: The x-t and v-t graphs show the motion of a car
starting from a stop sign. What is the car’s acceleration.
Solution: Acceleration is defined as the slope of the v-t graph.
The graph rises by 3 m/s during a time interval of 3 s, so the
acceleration is (3 m/s)/(3 s)=1 m/s
2
.
Incorrect solution #1: The final velocity is 3 m/s, and
acceleration is velocity divided by time, so the acceleration is (3
m/s)/(10 s)=0.3 m/s
2
.
#
The solution is incorrect because you can’t find the slope of a
graph from one point. This person was just using the point at the
right end of the v-t graph to try to find the slope of the curve.
Incorrect solution #2: Velocity is distance divided by time so
v=(4.5 m)/(3 s)=1.5 m/s. Acceleration is velocity divided by time,
so a=(1.5 m/s)/(3 s)=0.5 m/s
2
.
#
The solution is incorrect because velocity is the slope of the
tangent line. In a case like this where the velocity is changing,
you can’t just pick two points on the x-t graph and use them to
find the velocity.
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