A car is moving in a straight line along a horizontal road - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 2

Next, we compute the area of the trapezoid below the time axis from $10$ to $15$ seconds. The area can be expressed as:

$\text{Area}_{\text{below}} = 2(10 - t) + 20$

where $t$ is the time. We will evaluate this for the interval from $10$ to $15$ seconds.

The total displacement is the area above minus the area below. Thus, we set up the equation:

$\text{Total Displacement} = \text{Area}_{\text{above}} - \text{Area}_{\text{below}} = 20 - (2(10 - t) + 20)$

Setting this equal to $-7$ gives:

$20 - (20 - 2t) = -7$

Simplifying the equation:

$2t = 7 \implies t = 3.5 \, \text{seconds}$

We need to consider the period after $10$ seconds. The equation continues from there. By evaluating when the car returns to $0 \, \text{ms}^{-1}$ for the next instance:

From the graph, the velocity appears to hit zero again at: $t = 8.25 \, \text{seconds}$