Exercise solution 9, page 291 - AQA GCSE MATHS HIGHER

⭐ What you need to know:

A) On a distance–time graph, the horizontal axis shows time and the vertical axis shows distance. To read a point on the graph, match its time coordinate with its distance coordinate. For example, if a point is at \(4\) seconds and \(18\) metres, this means that after \(4\) seconds the runner has covered \(18\) metres.

B) The total distance run is the final distance shown on the graph. You look at the last point of the journey and read its value on the distance axis. This tells you how far the sprinter has travelled altogether.

C) Speed tells us how much distance is covered in a given time. The basic formula is:

\[ \text{speed}=\frac{\text{distance}}{\text{time}} \]

For example, if an athlete runs \(16\) metres in \(2\) seconds, then:

\[ \text{speed}=\frac{16}{2}=8\text{ m/s} \]

When the question asks for the speed in the first few seconds, use the distance covered during that time interval and divide by the number of seconds.

D) The average speed for a whole journey is found by dividing the total distance by the total time:

\[ \text{average speed}=\frac{\text{total distance}}{\text{total time}} \]

This gives the speed in metres per second if the distance is in metres and the time is in seconds.

E) To convert from m/s to km/h, multiply by \(3.6\):

\[ 1\text{ m/s}=3.6\text{ km/h} \]

For example, if a speed is \(7\text{ m/s}\), then:

\[ 7\times 3.6=25.2\text{ km/h} \]

This is useful when a question asks for average speed in km/h instead of m/s.

F) The gradient of a distance–time graph represents speed. A steeper graph means a greater speed. If the curve becomes steeper as time goes on, the runner is speeding up. If it becomes less steep, the runner is slowing down. A straight line means constant speed.

G) Acceleration is the rate at which speed changes. Deceleration means the speed is decreasing. If deceleration is constant, you can use:

\[ \text{deceleration}=\frac{\text{change in speed}}{\text{time taken}} \]

If something slows from \(12\text{ m/s}\) to \(0\text{ m/s}\) in \(4\) seconds, then:

\[\begin{aligned} \text{deceleration} &= \frac{12-0}{4} \\ &= \frac{12}{4} \\ &= 3\text{ m/s}^2 \end{aligned}\]

When an object comes to a complete stop, its final speed is \(0\text{ m/s}\).

H) Always check the units in your answer. Distance may be in metres, time in seconds, speed in m/s, and average speed may need converting to km/h. Writing the correct unit is an important part of a complete answer.