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# 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.

### 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".

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). ### Recall: Area of a trapezoid

NOTE: A trapezoid has one pair of parallel sides. (It's called a "trapezium" in British English.)

A trapezoid (trapezium), parallel sides a and b, height h.

We find the area of a trapezoid by simply finding the average of the two parallel sides, and multiplying by the distance between them (usually called the "height"):

text(Area)=((a+b)h)/2

We'll make use of this result in the next question.

### Example 2

A body moves as described by the following v-t graph. a) Describe the motion.

b) What is the distance travelled during the motion?

c) What is the average speed for the motion?

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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