A distance–time graph is plotted for a runner during a training session. From t = 0 to t = 8 s, the graph is a straight line and the distance increases steadily from 0 m to 96 m. From t = 8 s to t = 20 s, the graph is a horizontal straight line at a distance of 96 m. What is the…

IGCSE Physics 0625: Reading Speed and Rest from a Distance-Time Graph

Step 1: Calculate the speed from the first section of the graph

The first section is a straight line, so the runner’s speed is constant during this time. Speed is the gradient of a distance–time graph, found using:

$v = \frac{s}{t}$

Using the values for the first section only ( $8\text{ s}$ , not the whole $20\text{ s}$ ):

$v = \frac{96\text{ m}}{8\text{ s}} = 12\text{ m/s}$

Step 2: Interpret the second section of the graph

From $t = 8\text{ s}$ to $t = 20\text{ s}$ , the graph is horizontal: the distance stays at $96\text{ m}$ and does not increase. A horizontal (flat) section of a distance–time graph has a gradient of zero, which means the speed is zero. The runner is at rest.

Why the other options are wrong

Option	What it claims	Why it’s wrong
B	$12\text{ m/s}$ , constant speed	Misreads a flat line as continued motion, but a flat line means the distance is not changing at all
C	$4.8\text{ m/s}$ , at rest	Uses the total time ( $20\text{ s}$ ) instead of the time for just the first section ( $8\text{ s}$ )
D	$12\text{ m/s}$ , accelerating	A flat line has zero gradient (zero speed), not a changing gradient, so it cannot show acceleration

Final answers

Speed during the first $8\text{ s}$ $=$ $12\text{ m/s}$
During the next $12\text{ s}$ , the runner is at rest, option A