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

simplify “m/s/s” to read “m/s

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.