The diagram (not drawn to scale) shows the speed-time graph of a vehicle. The vehicle starts from rest and accelerates uniformly to a speed of 20 m/s in 15 seconds. It then travels at this speed before it decelerates at 1 m/s^2 until it comes to rest.

1. Find the uniform acceleration during the first 15 seconds.

Acceleration $=\dfrac{\text{change in speed}}{\text{time}}$

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

2. Find the speed after 10 seconds using the constant acceleration.

At $t=10$ s,

$v=0+\dfrac{4}{3}\times 10=\dfrac{40}{3}=13.333\ldots$

1. Find the time taken to decelerate from $20\text{ m/s}$ to rest.

Deceleration $=1\text{ m/s}^2$.

Time to decelerate
$=\frac{20}{1}=20\text{ s}.$

2. Express the total distance as the area under the speed-time graph.

Let the vehicle travel at $20\text{ m/s}$ from $t=15$ to $t=P-20$.

Total distance:
$\frac12(15)(20)+20[(P-20)-15]+\frac12(20)(20)=950.$

3. Solve the equation for $P$.

$150+20(P-35)+200=950$
$20(P-35)=600$
$P-35=30$
$P=65.$