The diagram show the speed-time graph of a car's journey between two road junctions.
The shaded area represents the distance travelled.
The distance travelled is 1620 m.

1. Relate the distance travelled to the triangular area under the speed-time graph.

Distance travelled $=$ area under speed-time graph.

$\frac{1}{2}\times 72\times v=1620$

2. Solve the equation to find the greatest speed.

$36v=1620$

$v=\frac{1620}{36}=45$

1. Use the proportional relationship on the straight-line part of the speed-time graph.

From part (a), the greatest speed is $45\text{ m/s}$ at $48$ seconds.

For the first $48$ seconds, the graph is a straight line from $(0,0)$ to $(48,45)$, so
$\frac{\text{speed at }32\text{ s}}{45}=\frac{32}{48}.$

2. Calculate the speed after $32$ seconds.

$\text{speed at }32\text{ s}=45\times \frac{32}{48}=30$

1. Calculate the decrease in speed per second over the last 24 seconds.

For the last $24$ seconds, the speed decreases from $45\text{ m/s}$ to $0\text{ m/s}$.

$\text{deceleration}=\frac{45-0}{24}=1.875\text{ m/s}^2$