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

Next, we calculate the area of the trapezium below the time axis from t = 10 to t = 15. The height forms a horizontal line at y = -3 m s⁻¹ over a base of 5 seconds with an additional triangular region between t = 10 to t = 15.

The area of the trapezium below the time axis can be calculated as follows: $\text{Area below} = \text{Area of rectangle} + \text{Area of triangle}$

Where:

• Area of rectangle (10 to 15 seconds, height = -3): $\text{Area} = \text{base} \times \text{height} = 5 \times -3 = -15\ \, \text{m}$
• Area of a triangle from 5 to 10 seconds: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times -3 = -7.5\ \, \text{m}$ Thus, the total area below the axis is: $\text{Total area below} = -15 - 7.5 = -22.5\ \, \text{m}$

The total displacement, D, is found from the areas calculated above.

Given that the total displacement over the period 0 ≤ t ≤ 15 is -7 m:

$D = \text{Area above} + \text{Area below} = 20 + (-22.5) + x = -7\,$ where x is the area yet to be considered.

Solving for x gives: $x = -7 - 20 + 22.5\ = -4.5\ \, \text{m}$

From the graph, the velocity becomes 0 m s⁻¹ at the initial instance at t = 0 seconds and also at a later point after t = 10 seconds. Since the velocity becomes positive again after t = 10 seconds, and decreases back to 0 at about t = 8.25 seconds, the next time when the velocity is 0 m s⁻¹ can be calculated from the equation setting the velocity equation equal to zero:

The solution yields: $t = 8.25 \, \text{seconds}$