上图显示了一个质点的速度-时间图像。

1. Identify the relevant straight-line section and find its gradient

For $20 \leq t \leq 35$, the graph is a straight line from $(20,10)$ to $(35,0)$.

Gradient $= \dfrac{0-10}{35-20}= -\dfrac{10}{15}= -\dfrac{2}{3}$ m/s per second.

2. Calculate the speed at $t=25$ using the constant rate of decrease

From $t=20$ to $t=25$, time elapsed $=5$ s.

Decrease in speed $= \dfrac{2}{3}\times 5=\dfrac{10}{3}$ m/s.

Speed at $t=25$:
$10-\dfrac{10}{3}=\dfrac{20}{3}\text{ m/s}$
$\dfrac{20}{3}=6.67\text{ m/s}\text{, to 3 s.f.}$

1. Identify and calculate the three areas under the speed-time graph

Distance covered in the first $35$ seconds = area under the speed-time graph.

From $0$ to $10$ s: triangle
$\frac12 \times 10 \times 10=50$

From $10$ to $20$ s: rectangle
$10\times 10=100$

From $20$ to $35$ s: triangle
$\frac12 \times 15 \times 10=75$

2. Add the component distances to obtain the total distance

Total distance
$=50+100+75=225\text{ m}$

1. Identify the key distance values to plot on the distance-time graph

From part (a)(ii), the total distance in the first $35$ seconds is $225\text{ m}$.

The intermediate distances are found from the areas under the speed-time graph:

At $t=10$,
$\text{distance}=\frac12(10)(10)=50\text{ m}.$

At $t=20$,
$\text{distance}=50+(10)(10)=150\text{ m}.$

So the key points are
$(0,0),\ (10,50),\ (20,150),\ (35,225).$

2. Draw the correct distance-time graph shape using the changing speed

Since the speed is the gradient of the distance-time graph:

• from $0$ to $10$ seconds, speed increases, so the graph rises with increasing gradient: draw a concave-up curve to $(10,50)$;
• from $10$ to $20$ seconds, speed is constant at $10\text{ m/s}$, so draw a straight line to $(20,150)$;
• from $20$ to $35$ seconds, speed decreases to $0$, so draw a rising concave-down curve to $(35,225)$, ending horizontal.