Graphs of Motion · Form 4

Graphs of Motion: Worked Examples (KBAT)

Q: What do KBAT questions on graphs of motion typically ask?

KBAT questions usually compare two moving objects on the same axes, say, two cyclists or a car and a bus, and ask when they meet, who is faster at a given time, or to sketch a missing part of the graph from a description. You need to interpret the story behind the graph, not just calculate gradients.

Q: How do I find where two objects meet on a combined distance-time graph?

The objects meet at the point where their two graph lines cross, since that is when both have travelled the same distance at the same time. Read the time and distance coordinates of that crossing point directly off the graph, or set the two distance expressions equal to each other if you have equations for both lines.

Q: What's a common pitfall when sketching a missing section of a motion graph from a description?

Students often draw a straight line even when the description implies changing speed, or forget that the sketch must connect smoothly to the given part of the graph at the same point. Re-read the description phrase by phrase, matching each part to a straight or curved section, and check your sketch starts exactly where the given graph ends.

These are real, multi-step situations where the graph technique is hidden, an LRT run between stations, a cycle trip with a shop stop, and a highway stopping-distance check. It helps students who can do single steps but freeze when the question is dressed up.

Written by the spmmath.com.my editorial team.

Worked example 1

An LRT train covers the 1500 m between two stations like this: from rest it accelerates uniformly to 25 m/s in 20 s, runs at a constant 25 m/s, then decelerates uniformly to rest in 15 s at the next station. Find how long it runs at the constant 25 m/s, and the total time between the two stations.

Sketch the speed-time graph: a triangle (speeding up), a rectangle (constant), then a triangle (slowing down); the whole area = 1500 m.
Distance while speeding up = ½ × 20 × 25 = 250 m.
Distance while slowing down = ½ × 15 × 25 = 187.5 m.
Distance at constant speed = 1500 − 250 − 187.5 = 1062.5 m.
Time at constant speed = distance ÷ speed = 1062.5 ÷ 25 = 42.5 s.
Total time = 20 + 42.5 + 15 = 77.5 s.

Worked example 2

Aisyah cycles 6 km to school. The first 4 km take 8 minutes at a steady speed; she then stops 4 minutes at a shop; the last 2 km take 8 minutes at a steady speed.

Find her speed on the faster moving section in km/h, and her average speed for the whole trip (including the stop) in km/h.

First section speed = distance ÷ time = 4 km ÷ (8/60 h) = 4 ÷ 0.1333… = 30 km/h.
Last section speed = 2 km ÷ (8/60 h) = 2 ÷ 0.1333… = 15 km/h.
The faster moving section is the first one, at 30 km/h.
Total distance = 6 km; total time = 8 + 4 + 8 = 20 minutes = 20/60 h = ⅓ h.
Average speed = 6 ÷ (1/3) = 18 km/h.

Worked example 3

A car travels at 30 m/s on a highway. The driver takes 0.8 s to react, keeping the speed at 30 m/s, then brakes uniformly to rest in 4 s.

A stalled lorry is 100 m ahead when the driver first sees it. Using a speed-time graph, find the total stopping distance and decide whether the car stops in time.

Reaction stage: the speed stays 30 m/s, so the graph is a rectangle; distance = 30 × 0.8 = 24 m.
Braking stage: the speed falls from 30 m/s to 0 in 4 s, a triangle; distance = ½ × 4 × 30 = 60 m.
Total stopping distance = 24 + 60 = 84 m.
Compare with the gap: 84 m < 100 m, so the car stops about 16 m before the lorry.

Worked example 4

Encik Rahman drives from town P to town Q, a distance of 18 km. For the first 12 km he drives at 72 km/h, then heavy traffic slows him to 24 km/h for the remaining 6 km.

Find (a) the total time for the journey in minutes, and (b) his average speed for the whole journey in km/h.

Time for first part = 12 ÷ 72 = 1/6 hour = 10 minutes.
Time for second part = 6 ÷ 24 = 1/4 hour = 15 minutes.
(a) Total time = 10 + 15 = 25 minutes = 25/60 = 5/12 hour.
(b) Average speed = total distance ÷ total time = 18 ÷ (5/12) = 18 × 12/5.
Average speed = 43.2 km/h.

Worked example 5

Puan Siti jogs from her home to the park, a distance of 3 km, at 6 km/h. She then walks back home along the same route at 4 km/h.

Find her average speed for the whole trip in km/h.

Time to the park = 3 ÷ 6 = 0.5 hour.
Time back home = 3 ÷ 4 = 0.75 hour.
Total distance = 3 + 3 = 6 km; total time = 0.5 + 0.75 = 1.25 hours.
Average speed = total distance ÷ total time = 6 ÷ 1.25.
Average speed = 4.8 km/h.

Worked example 6

Two cyclists start from the same point at the same instant along a straight path. Cyclist P moves at a constant speed of 8 m/s.

Cyclist Q starts from rest and accelerates uniformly at 1 m/s². Using areas under their speed-time graphs, find (a) the time when Q catches up with P and (b) the distance from the starting point at that moment.

Distance of P after t seconds = area of rectangle = 8 × t = 8t.
Q's speed after t seconds = 1 × t = t, so Q's distance = area of triangle = ½ × t × t = 0.5t².
Q catches P when the distances are equal: 0.5t² = 8t.
Divide both sides by t (t ≠ 0): 0.5t = 8, so t = 16 s.
(b) Distance = 8 × 16 = 128 m.

Worked example 7

A lorry travels from Town M to Town N, 210 km apart. It covers the first 90 km at a constant speed of 60 km/h, then continues at a different constant speed for the remaining 120 km.

If the whole journey takes 3 hours, find (a) the speed during the second part of the journey, and (b) the average speed for the whole journey.

(a) Time for first part = 90 ÷ 60 = 1.5 h
Time for second part = total time − time for first part = 3 − 1.5 = 1.5 h
Speed for second part = distance/time = 120 ÷ 1.5 = 80 km/h
(b) Average speed = total distance/total time = 210 ÷ 3 = 70 km/h

Worked example 8

Town P and Town Q are 150 km apart. A bus leaves P for Q at a constant speed of 50 km/h.

At the same instant, a car leaves Q for P along the same road at a constant speed of 70 km/h. Find (a) the time taken for the bus and the car to meet, and (b) the distance from P at which they meet.

(a) As they move towards each other, their speeds add: combined speed = 50 + 70 = 120 km/h
Time to meet = distance/combined speed = 150 ÷ 120 = 1.25 h (1 h 15 min)
(b) Distance from P = bus's speed × time = 50 × 1.25 = 62.5 km

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What do KBAT questions on graphs of motion typically ask?

KBAT questions usually compare two moving objects on the same axes, say, two cyclists or a car and a bus, and ask when they meet, who is faster at a given time, or to sketch a missing part of the graph from a description. You need to interpret the story behind the graph, not just calculate gradients.

How do I find where two objects meet on a combined distance-time graph?

The objects meet at the point where their two graph lines cross, since that is when both have travelled the same distance at the same time. Read the time and distance coordinates of that crossing point directly off the graph, or set the two distance expressions equal to each other if you have equations for both lines.

What's a common pitfall when sketching a missing section of a motion graph from a description?

Students often draw a straight line even when the description implies changing speed, or forget that the sketch must connect smoothly to the given part of the graph at the same point. Re-read the description phrase by phrase, matching each part to a straight or curved section, and check your sketch starts exactly where the given graph ends.