The diagram shows the speed-time graph for a cyclist's journey for a period of 90 seconds. The cyclist accelerates at $0.25\ \mathrm{m/s}^2$ in the first 24 seconds. He then travels at a constant speed of 8 m/s for a distance of 288 m.

1. Find the initial speed y using acceleration.

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

$0.25=\dfrac{8-y}{24}$

$8-y=0.25\times24=6$

$y=2$

2. Find x using the distance travelled at constant speed.

For the constant-speed section,

$\text{time}=\dfrac{\text{distance}}{\text{speed}}=\dfrac{288}{8}=36\text{ s}$

This section starts at $t=24$, so

$x=24+36=60$

1. Find the distance travelled by $t=24$ using the first part of the speed-time graph.

From part (a), $y=2$.

Distance travelled from $t=0$ to $t=24$ is the area of a trapezium under the speed-time graph:

$\text{distance}=\frac12(2+8)(24)=120\text{ m}.$

So the distance-time graph passes through

$(24,120).$

2. Find the distance at $t=60$ and at $t=90$.

From part (a), $x=60$.

The constant-speed section covers $288$ m, so the distance at $t=60$ is

$120+288=408\text{ m}.$

For the final section, the distance from $t=60$ to $t=90$ is the area of a triangle:

$\text{distance}=\frac12(30)(8)=120\text{ m}.$

Hence the total distance at $t=90$ is

$408+120=528\text{ m}.$

So the graph also passes through

$(60,408)\quad \text{and}\quad (90,528).$

3. Complete the distance-time graph with the correct shapes and gradients.

Plot the key points:

$(0,0),\ (24,120),\ (60,408),\ (90,528).$

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

• from $0$ to $24$ s, the speed increases, so draw a curve increasing and getting steeper;
• from $24$ to $60$ s, the speed is constant at $8$ m/s, so draw a straight line;
• from $60$ to $90$ s, the speed decreases to $0$, so draw a curve increasing but getting less steep, ending with gradient $0$ at $t=90$.