A train travels between two stations in a straight line. Its distance from the first station is recorded every so often, and the distance-time graph between each pair of readings is a straight line: Time, t / s 0 20 50 70 90 Distance, s / m 0 400 400 800 1000 (a) Calculate the t…

Answer: ((a)) 20 m/s ((b)) t = 20 s to t = 50 s; the distance stays at 400 m over this 30 s interval, so it is not changing, which means the train's speed is zero. ((c)) For t = 50 s to t = 70 s, speed = 20 m/s; for t = 70 s to t = 90 s, speed = 10 m/s ((d)) The train is slowing down (decelerating), because its speed falls from 20 m/s to 10 m/s between these two sections, as it approaches the second station. ((e)) 11.1 m/s (to 3 s.f.). Part (a): Speed during the first section Speed is found from the change in distance divided by the change in time, v = s/ t. Between t = 0 and t = 20 s: s = 400 m - 0 m = 400 m, t = 20 s - 0 s = 20 s v = 400 m/20 s = 20 m/s Part (b): Ident…

IGCSE Physics 0625: Speed from a Table of Distance-Time Readings

Time, $t$ / s	$0$	$20$	$50$	$70$	$90$
Distance, $s$ / m	$0$	$400$	$400$	$800$	$1000$

Time,

t

/ s

0

20

50

70

90

Distance,

s

/ m

0

400

400

800

1000

Part (a): Speed during the first section

Speed is found from the change in distance divided by the change in time, $v = \dfrac{\Delta s}{\Delta t}$ . Between $t = 0$ and $t = 20\text{ s}$ :

$\Delta s = 400\text{ m} - 0\text{ m} = 400\text{ m}, \qquad \Delta t = 20\text{ s} - 0\text{ s} = 20\text{ s}$

$v = \frac{400\text{ m}}{20\text{ s}} = 20\text{ m/s}$

Part (b): Identifying the section at rest

Comparing consecutive readings in the table, the distance is $400\text{ m}$ at both $t = 20\text{ s}$ and $t = 50\text{ s}$ . It has not changed over this $30\text{ s}$ interval. Since speed is the change in distance divided by the change in time, and the change in distance here is zero, the train’s speed during this section is zero: the train is at rest from $t = 20\text{ s}$ to $t = 50\text{ s}$ .

Part (c): Speed during the next two sections

From $t = 50\text{ s}$ to $t = 70\text{ s}$ :

$\Delta s = 800\text{ m} - 400\text{ m} = 400\text{ m}, \qquad \Delta t = 70\text{ s} - 50\text{ s} = 20\text{ s}$

$v = \frac{400\text{ m}}{20\text{ s}} = 20\text{ m/s}$

From $t = 70\text{ s}$ to $t = 90\text{ s}$ :

$\Delta s = 1000\text{ m} - 800\text{ m} = 200\text{ m}, \qquad \Delta t = 90\text{ s} - 70\text{ s} = 20\text{ s}$

$v = \frac{200\text{ m}}{20\text{ s}} = 10\text{ m/s}$

Part (d): Speeding up or slowing down

The train’s speed falls from $20\text{ m/s}$ (between $t = 50\text{ s}$ and $t = 70\text{ s}$ ) to $10\text{ m/s}$ (between $t = 70\text{ s}$ and $t = 90\text{ s}$ ). Since its speed is decreasing, the train is slowing down (decelerating). This is consistent with it approaching and preparing to stop at the second station.

Part (e): Average speed for the whole journey

Average speed uses the total distance travelled and the total time taken, not the individual section speeds:

$\text{total distance} = 1000\text{ m} - 0\text{ m} = 1000\text{ m}, \qquad \text{total time} = 90\text{ s} - 0\text{ s} = 90\text{ s}$

$\text{average speed} = \frac{1000\text{ m}}{90\text{ s}} = 11.111...\text{ m/s} \approx 11.1\text{ m/s}$

Final answers

(a) Speed from $t = 0$ to $t = 20\text{ s}$ $=$ $20\text{ m/s}$
(b) The train is at rest between $t = 20\text{ s}$ and $t = 50\text{ s}$
(c) Speed from $t = 50\text{ s}$ to $t = 70\text{ s}$ $=$ $20\text{ m/s}$ ; speed from $t = 70\text{ s}$ to $t = 90\text{ s}$ $=$ $10\text{ m/s}$
(d) The train is slowing down (decelerating) as it nears the second station
(e) Average speed for the whole $90\text{ s}$ journey $=$ $11.1\text{ m/s}$