下图显示了一个物体在 40 秒时间内的速度-时间图像。

1. Find the gradient of the decreasing speed-time graph section containing $t=35$.

At $t=35$, the object is on the straight-line section from $(28,23)$ to $(40,0)$.

$\text{acceleration}=\frac{0-23}{40-28}=-\frac{23}{12}\text{ m/s}^2$

Hence the deceleration is

$\frac{23}{12}\text{ m/s}^2.$

1. Find the acceleration on the straight-line section from $t=10$ to $t=28$.

From $t=10$ to $t=28$, the speed increases linearly from $11$ m/s to $23$ m/s.

$\text{acceleration}=\frac{23-11}{28-10}=\frac{12}{18}=\frac{2}{3}\text{ m/s}^2$

2. Use the acceleration to find the speed at $t=18$.

At $t=18$, the time elapsed since $t=10$ is

$18-10=8\text{ s}$

Increase in speed:

$\frac{2}{3}\times 8=\frac{16}{3}\text{ m/s}$

Speed at $t=18$:

$11+\frac{16}{3}=\frac{49}{3}=16\frac{1}{3}\text{ m/s}$

1. Find the distance travelled by $t=10$ from the rectangular area under the speed-time graph.

From $t=0$ to $t=10$, speed is constant at $11\text{ m/s}$.

$\text{Distance}=11\times 10=110\text{ m}$

2. Find the cumulative distance travelled by $t=28$ using the trapezium area from $t=10$ to $t=28$.

From $t=10$ to $t=28$, the area under the speed-time graph is a trapezium with parallel sides $11$ and $23$, and width $18$.

$\text{Distance from }10\text{ to }28=\frac12(11+23)(18)=306\text{ m}$

$\text{Distance at }t=28=110+306=416\text{ m}$

3. Find the total distance by $t=40$ and describe the required shape of the distance-time graph.

From $t=28$ to $t=40$, the area under the speed-time graph is a triangle.

$\text{Distance from }28\text{ to }40=\frac12(12)(23)=138\text{ m}$

$\text{Distance at }t=40=416+138=554\text{ m}$

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

• from $0$ to $10$, sketch a straight line with constant positive gradient;
• from $10$ to $28$, sketch a curve getting steeper;
• from $28$ to $40$, sketch a curve getting less steep, ending horizontal at $t=40$.

The graph should pass through $(0,0)$, $(10,110)$, $(28,416)$ and $(40,554)$.