The diagrams show the speed-time graph and the corresponding distance-time graph of a car. The car travelled from a point P to Q and in the journey, its greatest speed attained was 1.5 km/min.

1. Convert $1.5\text{ km/min}$ to metres per second.

$1.5\text{ km/min}=\dfrac{1.5\times1000}{60}\text{ m/s}=25\text{ m/s}$

1. Relate the distance travelled from 0800 to 0820 to the area under the speed-time graph.

From 0800 to 0820, time taken $=20$ min.

The distance travelled is $21$ km, so
$\text{Area under speed-time graph}=21.$
Since the graph is a trapezium,
$\frac{1}{2}(v+1.5)(20)=21.$

2. Solve the resulting linear equation for the initial speed $v$.

$\frac{1}{2}(v+1.5)(20)=21$
$10(v+1.5)=21$
$v+1.5=2.1$
$v=0.6$

1. Find the speed at 0812 using the linear increase from 0800 to 0820.

From part (b), the speed at 0800 was $0.6\text{ km/min}$.

From 0800 to 0820, the speed increases uniformly from $0.6$ to $1.5\text{ km/min}$.

0812 is $12$ minutes after 0800, so
$\text{speed at 0812}=0.6+\frac{12}{20}(1.5-0.6).$
$=0.6+\frac{12}{20}(0.9)=1.14\text{ km/min}$

2. Convert the speed from km/min to m/s.

$1.14\text{ km/min}=1.14\times \frac{1000}{60}\text{ m/s}$
$=19\text{ m/s}$

1. Find the distance travelled by 0900 and hence the remaining distance after 0900.

From 0820 to 0900, the car travels at $1.5\text{ km/min}$ for $40$ minutes.

Distance from 0820 to 0900:
$1.5\times 40=60\text{ km}$

Distance from 0800 to 0820 is $21\text{ km}$.

Distance travelled by 0900:
$21+60=81\text{ km}$

Remaining distance after 0900:
$92.25-81=11.25\text{ km}$

2. Use the triangular area under the deceleration part of the speed-time graph to find when the car stops.

After 0900, the speed decreases linearly from $1.5\text{ km/min}$ to $0$, so the area under the speed-time graph is a triangle.

Let the time taken after 0900 be $t$ minutes.

$\frac12\times t\times 1.5=11.25$

$0.75t=11.25$

$t=15$

So the car stops $15$ minutes after 0900.

$T=0915$