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Graphs of Motion · Form 4

Graphs of Motion: Worked Examples (Medium)

This set joins two or three steps: a distance-time journey that includes a rest, a trapezium area for total distance, and finding a deceleration together with its distance. It helps students moving from single readings to combined problems.

Written by the spmmath.com.my editorial team.

Worked example 1

A car's distance-time graph has three straight sections: from O it travels 90 km in 1.5 hours, then rests (a horizontal line) for 0.5 hour, then travels a further 30 km in 1 hour. Find (a) the speed during the last section, and (b) the average speed for the whole journey.

  1. (a) Speed of the last section = gradient = distance ÷ time = 30 ÷ 1 = 30 km/h.
  2. (b) Average speed uses total distance ÷ total time, and the rest time must be included.
  3. Total distance = 90 + 30 = 120 km.
  4. Total time = 1.5 + 0.5 + 1 = 3 hours.
  5. Average speed = 120 ÷ 3 = 40 km/h.

Worked example 2

On a speed-time graph a train speeds up uniformly from 8 m/s to 20 m/s in 6 s, then keeps a constant 20 m/s for 10 s. Find the total distance travelled in the 16 s.

  1. Stage 1 (speeding up) is a trapezium; distance = area = ½ × (sum of the two parallel speeds) × time = ½ × (8 + 20) × 6.
  2. = ½ × 28 × 6 = 84 m.
  3. Stage 2 (constant speed) is a rectangle; distance = 20 × 10 = 200 m.
  4. Total distance = 84 + 200 = 284 m.

Worked example 3

A bus moving at 24 m/s brakes uniformly and stops in 8 s. From its speed-time graph, find (a) the deceleration, and (b) the distance it covers while stopping.

  1. (a) Gradient = (final speed − initial speed) ÷ time = (0 − 24) ÷ 8 = −3 m/s²; the negative sign shows slowing down, so the deceleration is 3 m/s².
  2. (b) From 24 m/s to 0 over 8 s the shape is a triangle; distance = area = ½ × base × height.
  3. = ½ × 8 × 24 = 96 m.

Worked example 4

A distance-time graph rises as a straight line from O to (8, 160), then stays horizontal from (8, 160) to (20, 160). Distance is in metres and time in seconds.

Find (a) the speed during the first 8 seconds, and (b) the average speed for the whole 20 seconds.

  1. (a) Speed in first section = gradient = 160 ÷ 8 = 20 m/s.
  2. The horizontal part means the object is at rest, so total distance = 160 m.
  3. (b) Average speed = total distance ÷ total time = 160 ÷ 20.
  4. Average speed = 8 m/s.

Worked example 5

On a speed-time graph a train's speed falls uniformly from 30 m/s to 10 m/s in 8 seconds, then stays constant at 10 m/s for the next 12 seconds. Find the total distance travelled in the 20 seconds.

  1. First section is a trapezium: area = ½ × (30 + 10) × 8 = ½ × 40 × 8 = 160 m.
  2. Second section is a rectangle: area = 10 × 12 = 120 m.
  3. Total distance = 160 + 120.
  4. Total distance = 280 m.

Worked example 6

A van's speed increases uniformly from 12 m/s to 32 m/s over 5 seconds, shown as a straight line on a speed-time graph. Find (a) the acceleration and (b) the distance travelled during these 5 seconds.

  1. (a) Acceleration = (32 − 12) ÷ 5 = 20 ÷ 5 = 4 m/s².
  2. (b) Distance = area of trapezium = ½ × (12 + 32) × 5.
  3. Distance = ½ × 44 × 5 = 110 m.

Worked example 7

A car's speed-time graph has three straight sections: it accelerates uniformly from rest to 20 m/s in 5 s, travels at a constant 20 m/s for the next 15 s, then decelerates uniformly to rest in 4 s. Find the total distance travelled.

  1. Distance = total area under the speed-time graph, in three parts
  2. Section 1 (triangle): ½ × 5 × 20 = 50 m
  3. Section 2 (rectangle): 15 × 20 = 300 m
  4. Section 3 (triangle): ½ × 4 × 20 = 40 m
  5. Total distance = 50 + 300 + 40 = 390 m

Worked example 8

A cyclist's speed-time graph shows the speed decreasing uniformly from 12 m/s to 4 m/s in the first 8 s, then increasing uniformly from 4 m/s back to 12 m/s in the next 6 s. Find (a) the acceleration during the first 8 s, and (b) the total distance travelled in the 14 s.

  1. (a) Acceleration = (final speed − initial speed)/time = (4 − 12) ÷ 8 = −1 m/s² (a deceleration of 1 m/s²)
  2. (b) Distance in first 8 s (trapezium) = ½ × (12 + 4) × 8 = 64 m
  3. Distance in next 6 s (trapezium) = ½ × (4 + 12) × 6 = 48 m
  4. Total distance = 64 + 48 = 112 m

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

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Frequently asked questions

What's the key technique for medium-level graphs-of-motion questions?

Medium questions often involve a multi-stage journey, acceleration, constant speed, then deceleration, shown on one speed-time graph. The area under each section gives the distance for that stage, so split the graph into triangles and trapeziums, calculate each area, and add them for total distance.

Why is average speed not simply the average of the speeds shown on the graph?

Average speed is defined as total distance divided by total time, and each stage of a journey usually lasts a different length of time. Averaging the speed values directly ignores how long the object travelled at each speed, giving a wrong answer, always calculate total distance (area) and total time separately first.

How do I handle a graph question that gives some values as unknowns (like time = t)?

Write the area or gradient expression in terms of the unknown exactly as you would with a number, then use any given total (such as total distance or total time) to form an equation and solve for the unknown. Substitute your answer back into the original expression to find what the question actually asks for.

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