IGCSE · Practice questions · Conversion, distance-time and speed-time graphs

Gradient and area.

Six original Cambridge-style questions on graphs in practical situations: conversion graphs, distance-time and speed-time graphs, rates of change and tangents.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Check the axes, then gradient or area.

Axes: read the quantity, the units and the scale on each axis.
Gradient: vertical change ÷ horizontal change, with units (km/h, m/s², $ per £).
Units: change minutes to hours before working out km/h.
Area (Extended): split the area under a speed-time graph into triangles, rectangles and trapezia.
Curves (Extended): draw a tangent and use two far-apart points on it.

Recall

[3 marks]

A conversion graph for pounds (£) and dollars ($) is a straight line through $(0, 0)$ and (£80, $100).

(a) Convert £24 to dollars. (b) Convert $65 to pounds. (c) What does the gradient represent?

Gradient $= 100 \div 80 = 1.25$, so $ = 1.25 × £. ✓
(a) $24 \times 1.25 = 30$, so $30 ✓
(b) $65 \div 1.25 = 52$, so £52 ✓
(c) The exchange rate: 1.25 dollars for each pound. ✓

Recall

[2 marks]

Describe the motion shown on a distance-time graph by (a) a horizontal line, (b) a straight line steeper than the one before it.

(a) Stationary (not moving). ✓
(b) Moving at a constant, faster speed. ✓

Calculation

[2 marks]

A distance-time graph shows Mia walking 2.4 km in 36 minutes at a steady speed. Find her speed in km/h.

36 min $= \dfrac{36}{60} = 0.6$ h. ✓
Speed $= 2.4 \div 0.6 = 4$ km/h. ✓

Calculation

[3 marks]

(Extended) A car accelerates uniformly from 5 m/s to 25 m/s in 8 seconds. Find (a) its acceleration, (b) the distance it travels in these 8 seconds.

(a) $\dfrac{25 - 5}{8} = 2.5$ m/s². ✓
(b) Trapezium area $= \dfrac{5 + 25}{2} \times 8 = 120$ m. ✓

Analysis

[3 marks]

(Extended) An object moves so that its distance after $t$ seconds is $d = t^2$ metres. A tangent to the distance-time curve at $t = 3$ passes through $(1.5, 0)$ and $(5, 21)$.

(a) Estimate the speed at $t = 3$. (b) Find the average speed for the first 3 seconds and explain why it is different.

(a) Gradient of tangent $= \dfrac{21 - 0}{5 - 1.5} = \dfrac{21}{3.5} = 6$ m/s. ✓
(b) At $t = 3$, $d = 9$, so the average speed is $9 \div 3 = 3$ m/s. ✓
The object is speeding up, so its speed at $t = 3$ is greater than its average over the first 3 seconds. ✓

Analysis

[5 marks]

Leila leaves home at 09 00. She drives at 60 km/h for 45 minutes, stops for 30 minutes, then drives a further 30 km in 20 minutes.

(a) How far from home is she after the first stage? (b) At what time does she arrive? (c) Find her speed on the last stage. (d) Find her average speed for the whole journey. (e) Describe how the distance-time graph of the journey would look.

(a) $60 \times 0.75 = 45$ km. ✓
(b) $45 + 30 + 20 = 95$ min after 09 00, so 10 35. ✓
(c) $30 \div \dfrac{20}{60} = 90$ km/h. ✓
(d) Total 75 km in $\dfrac{95}{60}$ h: $75 \div \dfrac{95}{60} = 47.4$ km/h (3 s.f.). ✓
(e) A straight line from (09 00, 0) to (09 45, 45), horizontal to (10 15, 45), then a steeper line to (10 35, 75). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.