1.3 Acceleration
Acceleration is the rate at which the velocity changes.
Acceleration is represented by the gradient of a velocity–time graph.
It is a vector and can take different signs in a similar way to velocity. This is illustrated by Tom’s cycle journey which is shown in Figure 1.11.
Figure 1.11
Tom turns on to the main road at 4 m s−1, accelerates uniformly, maintains a constant speed and
then slows down uniformly to stop when he reaches home.
Between A and B, Tom’s velocity increases by (10 − 4) = 6 m s−1 in 6 seconds, that is by 1 meter per second every second.
This acceleration is written as 1 m s−2 (one meter per second squared) and is the gradient of AB.
From B to C: 
From C to D:
From C to D, Tom is slowing down while still moving in the positive direction towards home, so his acceleration, the gradient of the graph, is negative.
The sign of acceleration
Think again about the marble thrown up into the air with a speed of 5 m s−1 . Figure 1.12 represents the velocity when upwards is taken as the positive direction and shows that the velocity decreases from + 5 m s−1 to 5 m s−1 in 1 second. This means that the gradient, and hence the acceleration, is negative. It is −10 m s−2.
Figure 1.12
Question 1.3.1:
A lift travels up and down between the ground floor (G) and the roof garden (R) of a hotel. It starts from rest, takes 5 s to increase its speed uniformly to 2 m s−1, maintains this speed for 5 s and then slows down uniformly to rest in another 5 s. In the following questions, use upwards as positive.
(i) Sketch a velocity–time graph for the journey from G to R.
On one occasion the lift stops for 5 s at R before returning to G.
(ii) Sketch a velocity–time graph for this journey from G to R and back.
(iii) Calculate the acceleration for each 5 s interval. Take care with the signs.
(iv) Sketch an acceleration–time graph for this journey.
Solution 1.3.1:
(i)
(ii)
(iii)
Acceleration at each 5 s interval: +0.4 m s−2 , 0 m s−2, −0.4 m s−2, 0 m s−2 , −0.4 m s−2, 0 m s−2, +0.4 m s−2.
(iv)
