The graph below shows the velocity of an object moving in a straight line over a 20 second journey - AQA - A-Level Maths Pure - Question 12 - 2018 - Paper 2

To calculate the acceleration, we can determine the steepest gradient of the velocity-time graph. The acceleration is given by:

$a = \frac{\Delta v}{\Delta t}$

From the graph:

• The steepest increase occurs from 0 m/s to 3 m/s between 0s and 5s, giving an acceleration of:

$a = \frac{3 - 0}{5 - 0} = \frac{3}{5} = 0.6 \text{ m/s}^2$

• The steepest decrease is from 3 m/s to -4 m/s between 5s and 8s:

$a = \frac{-4 - 3}{8 - 5} = \frac{-7}{3} = -2.33 \text{ m/s}^2$

Taking the absolute value:

$|a| = 2.33 \text{ m/s}^2$

However, the most significant gradient happens from 12s to 14s where it goes from 1 m/s to -4 m/s:

$a = \frac{-4 - 1}{14 - 12} = \frac{-5}{2} = -2.5 \text{ m/s}^2$

From the graph, the maximum magnitude of acceleration is determined from these calculations, hence the answer is:

Maximum acceleration = 4 m/s².