The diagram below shows the speed-time graph of a car which starts from rest.

1. Find the constant rate of increase of speed from 0 to 15 seconds.

From $t=0$ to $t=15$, the graph is a straight line from $0\text{ m/s}$ to $20\text{ m/s}$.

Gradient $=\dfrac{20-0}{15-0}=\dfrac{20}{15}=\dfrac{4}{3}\text{ m/s}^2$.

2. Use the gradient to find the speed at 9 seconds.

Speed after $9$ seconds
$=\dfrac{4}{3}\times 9=12\text{ m/s}.$

1. Express the distance travelled in the first 40 seconds as areas under the speed-time graph.

Distance in first 40 s
$=\frac{1}{2}(15)(20)+\frac{1}{2}(20+u)(40-15).$
Since the distance is 900 m,
$\frac{1}{2}(15)(20)+\frac{1}{2}(20+u)(25)=900.$

2. Solve the resulting linear equation for $u$.

$150+12.5(20+u)=900$
$150+250+12.5u=900$
$12.5u=500$
$u=40.$

1. Find the magnitude of the gradient of the speed-time graph over the last 20 seconds.

From part (b)(i), $u=36$. Over the last $20$ s, the speed decreases from $36\text{ m/s}$ to $0\text{ m/s}$.

$\text{deceleration}=\frac{36-0}{60-40}=\frac{36}{20}=1.8\text{ m/s}^2$