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

To determine when the velocity becomes $0 \mathrm{ms^{-1}}$ again, we first need to analyze the graph.

The graph indicates that the car's velocity becomes zero when it comes to a stop. Observing the graph, we see that the velocity transitions to $0 \mathrm{ms^{-1}}$ again after some time.

We denote the time where the velocity is $0$ again as $t = t_{0}$. We can establish the area under the velocity-time graph to solve for displacement, using the concept that displacement is the net area under the velocity-time graph.

The area above the $t$-axis (corresponding to positive velocity) needs to be calculated, which forms a triangle above the $t$-axis.

1. Area above the time axis:

   • Base = $10$ seconds, Height = $4 \mathrm{ms^{-1}}$, so:
     $\text{Area above} = \frac{1}{2} \times 10 \times 4 = 20 \mathrm{m}$

2. Area below the time axis (where the car is reversing):

   • The triangle below from $10$ seconds to $15$ seconds has the base = $5$, Height = $3 \mathrm{ms^{-1}}$, so:
     $\text{Area below} = \frac{1}{2} \times 5 \times 3 = 7.5 \mathrm{m}$

Adding displacements, we have: $\text{Total displacement} = 20 - 7.5 = 12.5 \mathrm{m}$

Three areas can make up the total displacement of $-7$ m, hence we can set up an equation considering the additional area from $15$ seconds to $t_{0}$: $\text{Total area} = \text{Area above} - \text{Area below} - \text{Additional area} = -7 \mathrm{m}$

From previous calculations, we have: $20 - 7.5 - k = -7\Rightarrow k = 20 - 7.5 + 7 = 19.5 \mathrm{m}$

Given that additional displacement came from zero velocity which takes place until $t_{0}$ denoted below:
$t_{0} = 15 + t_{0} \Rightarrow 2(10 - t_{0}) + 19.5 = 0.5\Rightarrow 25 = k + 15 + t_{0} = 25\Rightarrow t_{0} = 8.25 \mathrm{seconds}$.

Thus, the next time the car has a velocity of $0 \mathrm{ms^{-1}}$ occurs at:
$t = 8.25 \mathrm{seconds}$.