The diagram shows the speed-time graph of two cars, A and B. Car A, starting from rest, accelerates uniformly for 2 seconds until it reaches a speed of 11 m/s. It then continues to travel at constant speed. 2 seconds later, Car B starts from the same point as Car A.

1. Find the acceleration from the gradient of Car A's speed-time graph

For Car A from $t=0$ to $t=2$,
$\text{acceleration}=\frac{\text{change in speed}}{\text{change in time}}=\frac{11-0}{2-0}=5.5\text{ m/s}^2.$
Since the acceleration is uniform, this is also the acceleration when $t=1$.

1. Find the distance from the area under Car A's speed-time graph for the first 2 seconds

Distance travelled $=$ area under speed-time graph

$= \dfrac{1}{2} \times 2 \times 11$

$= 11\text{ m}$

1. Find the total distance travelled by Car A from $t=0$ to $t=9$

Distance travelled by Car A is the area under its speed-time graph.

For the first $2$ seconds, distance $=11\text{ m}$ from part (a)(ii).

From $t=2$ to $t=9$, Car A travels at $11\text{ m/s}$ for $7$ seconds.

$\text{Distance by A}=11+11(7)=88\text{ m}$

2. Equate Car B's distance to Car A's distance and solve for $v$

Car B starts from rest at $t=2$ and overtakes Car A at $t=9$, so it travels for $7$ seconds.

Its speed-time graph is a triangle with base $7$ and height $v$.

Since Car B overtakes Car A, distances are equal:

$\frac12(7)(v)=88$

$v=\frac{176}{7}=25.142\ldots$

$v=25.1\text{ m/s}\text{, to 3 s.f.}$