2. Acceleration (`v`-`t`) Graphs

by M. Bourne

Acceleration is the change in velocity per time.

A common unit for acceleration is `"ms"^-2`. An acceleration of `7\ "ms"^-2` means that in each second, the velocity increases by `7\ "ms"^-1` (also written as `7\ "m/s"`).

We can find the acceleration by using the expression:

`text(acceleration)=text(change in velocity)/text(change in time`

We can write the above using the equivalent

`text(acceleration)=(Deltav)/(Deltat`

where the Greek letter `Δ` (Delta) means "change in".

In other words, the slope of the velocity graph tells us the acceleration.

The Area Under the `v`-`t` Graph

A very useful aspect of these graphs is that the area under the v-t graph tells us the distance travelled during the motion.

This concept is important when we find areas under curves later in the integration chapter.

Continues below

Example 1

A particle in a generator is accelerated from rest at the rate of `55\ "ms"^-2`.

a. What is the velocity at `t = 3\ "s"`?

b. What is the acceleration at `t = 3\ "s"`?

c. What is the distance travelled in `3` seconds?

d. Graph the acceleration (as a v - t graph) for `0 ≤ t ≤ 3\ "s"`.

Answer

a. Velocity `= 55 × 3 = 165\ "ms"^-1`

b. The acceleration is a constant `55\ "ms"^-2`, so at `t = 3\ "s"`, the acceleration will be `55\ "ms"^-2`.

c. The distance travelled in `3` seconds is `165 × 1.5 = 247.5\ "m"`. We obtain this from the area under the line between `0` and `3` (i.e. the area of the shaded triangle below).

d. Note in the graph that we have velocity on the vertical axis, and the units are m/s.

The graph finishes at (3, 165).

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Example 2

A body moves as described by the following v-t graph.

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a) Describe the motion.

b) What is the distance travelled during the motion?

c) What is the average speed for the motion?

Answer

a) From `t = 0` to `2`, the acceleration was `a=(Deltav)/(Deltat)=3/2=1.5\ text(ms)^-2`

From `t = 2` to `5`, the acceleration was `0\ "ms"^-1`.

The body was neither speeding up nor slowing down.

From `t = 5` to `8`, the acceleration was `a=(Deltav)/(Deltat)=(-3)/3=-1\ text(ms)^-2`

The body was slowing down, so the acceleration was negative.


b) The distance travelled is the area of the trapezoid (trapezium).

`text(distance)=((a+b)h)/2`

`=((8+3)(3))/2`

`=16.5\ text(m)`


c) `text(average speed)=text(distance travelled)/text(time taken)`

`=16.5/8`

`=2.1\ text(ms)^-1`

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