Motion and Motion Graphs: Question 6

Worked solution

Part (a): Speed during the outward section

Speed is the gradient of a distance–time graph, $v = \dfrac{s}{t}$. During the first section, the cyclist’s distance from home increases from $0\text{ m}$ to $1500\text{ m}$ over $300\text{ s}$:

$v = \frac{1500\text{ m}}{300\text{ s}} = 5.0\text{ m/s}$

Part (b): The resting section

Between $t = 300\text{ s}$ and $t = 420\text{ s}$, the cyclist’s distance from home does not change. It stays at $1500\text{ m}$. On the graph this is a horizontal (flat) line, which has a gradient of zero. Since the gradient of a distance–time graph gives the speed, a gradient of zero means the cyclist is at rest (she is resting at the park).

Part (c): Speed during the return section

Between $t = 420\text{ s}$ and $t = 570\text{ s}$, her distance from home decreases from $1500\text{ m}$ to $0\text{ m}$. The size of this change in distance is:

$\Delta s = 1500\text{ m} - 0\text{ m} = 1500\text{ m}$

and the time taken is:

$\Delta t = 570\text{ s} - 420\text{ s} = 150\text{ s}$

So her speed on the way home is:

$v = \frac{1500\text{ m}}{150\text{ s}} = 10\text{ m/s}$

Part (d): Why this shows she is travelling towards home

A distance-time graph always plots distance from a fixed point (here, home) against time. During the final section, this distance is getting smaller as time passes, the line slopes downward. This can only happen if the cyclist is moving closer to home, which is why this section represents the return journey rather than a continuation of the outward trip (which would show the distance still increasing).

Final answers

• (a) Speed on the way to the park $=$ $5.0\text{ m/s}$
• (b) The cyclist is at rest (resting at the park); the graph is horizontal, showing zero speed
• (c) Speed on the way home $=$ $10\text{ m/s}$
• (d) Her distance from home is decreasing, which shows she is travelling back towards home