Graphs of Motion · Form 4
Graphs of Motion: Worked Examples (Easier)
This set covers the three core reads of the chapter: turning the gradient of a straight distance-time graph into speed, the gradient of a speed-time graph into acceleration, and the area under a speed-time graph into distance. It helps students who are just starting graphs of motion.
Worked example 1
The distance-time graph of a jogger is a straight line from the origin O to the point (50, 150), where distance is in metres (m) and time is in seconds (s). Find the jogger's speed.
- On a distance-time graph, speed = gradient of the line.
- Gradient = change in distance ÷ change in time = (150 − 0) ÷ (50 − 0).
- = 150 ÷ 50 = 3 m/s.
Worked example 2
On a speed-time graph, a motorcycle's speed increases uniformly along a straight line from 5 m/s to 25 m/s over 8 s. Find its acceleration.
- On a speed-time graph, acceleration = gradient of the line.
- Gradient = (final speed − initial speed) ÷ time = (25 − 5) ÷ 8.
- = 20 ÷ 8 = 2.5 m/s².
Worked example 3
A train travels at a constant speed of 30 m/s for 40 s. Using its speed-time graph, find the distance travelled.
- At constant speed the graph is a horizontal line, so distance = area under the graph, a rectangle.
- Area = speed × time = 30 × 40.
- = 1200 m.
Worked example 4
A car travels along a straight road shown as a straight line on a distance-time graph from the origin O to the point (30, 240), where distance is in metres and time in seconds. Find the speed of the car.
- Speed = gradient of the distance-time graph = change in distance ÷ change in time.
- Speed = 240 m ÷ 30 s.
- Speed = 8 m/s.
Worked example 5
On a speed-time graph, a motorcycle's speed decreases uniformly from 18 m/s to 6 m/s in 4 seconds. Find the acceleration of the motorcycle.
- Acceleration = gradient = (final speed − initial speed) ÷ time.
- Acceleration = (6 − 18) ÷ 4.
- Acceleration = −12 ÷ 4 = −3 m/s² (a deceleration of 3 m/s²).
Worked example 6
A lorry moves at a constant speed of 15 m/s for 12 seconds. Using the area under the speed-time graph, find the distance travelled.
- The graph is a horizontal line, so the area is a rectangle.
- Distance = area = speed × time = 15 × 12.
- Distance = 180 m.
Worked example 7
A bus's distance-time graph is a straight line from the origin O to the point (2.5, 100), where distance is in kilometres (km) and time is in hours (h). Find the speed of the bus.
- Speed = gradient of the distance-time graph = distance/time
- Speed = 100 km ÷ 2.5 h
- Speed = 40 km/h
Worked example 8
A robot lawn mower moves in a straight line at a constant speed of 0.8 m/s for 50 s, shown as a horizontal line on its speed-time graph. Find the distance travelled, using the area under the graph.
- Distance = area under the speed-time graph = area of rectangle
- Distance = speed × time = 0.8 × 50
- Distance = 40 m
Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)
Frequently asked questions
What should I focus on for easy questions on graphs of motion?
Easy questions test basic reading skills: finding the gradient of a distance-time graph to get speed, or the gradient of a speed-time graph to get acceleration. Practise reading values off the axes accurately and remember that a flat section means no movement on a distance-time graph, but constant speed on a speed-time graph.
Why does units matter so much when calculating gradient in these questions?
The gradient's units depend entirely on the axes labels, a distance-time graph in km against hours gives speed in km/h, while minutes would give a different number for the same motion. Always check both axis labels before dividing, and write the correct unit with your final answer; a right number with the wrong unit still loses the mark.
What's a quick way to check if I've read the graph correctly before calculating?
Describe the motion in one sentence first, for example, "the object speeds up, then travels at a steady rate, then stops" before you calculate anything. If your calculated answer contradicts that description, such as a negative speed during a rising section, you've likely misread a coordinate or a scale.