速度-时间图象显示了林（Lim）在三小时行程中的旅程。

1. Find the gradient of the speed-time graph segment containing 0815

From 0800 to 0830, the speed decreases from $100$ km/h to $60$ km/h.

Time taken $=30$ min $=0.5$ h.

$\text{acceleration}=\frac{\text{change in speed}}{\text{change in time}}=\frac{60-100}{0.5}=-80\text{ km/h}^2$

1. Find the distance travelled from 0800 to 0830 to get $p$

From 0800 to 0830, the time taken is $30\text{ min}=0.5\text{ h}$.\n\nThe distance is the area of the trapezium under the speed-time graph:\n$p=\frac12(100+60)(0.5)=40$

2. Add the distance travelled from 0830 to 0930 to get $q$

From 0830 to 0930, the speed is constant at $60\text{ km/h}$ for $1\text{ h}$.\n\nDistance travelled in this interval:\n$60\times 1=60\text{ km}$\n\nHence\n$q=p+60=40+60=100$