Home · IGCSE 0625 · Motion, forces, energy · Acceleration and gradients · Practice
Extended · Practice questions · Acceleration and gradients

Subtract first, then divide.

Six original Cambridge-style Extended questions on defining acceleration, the change-in-velocity calculation, deceleration and its sign, and the v / t versus Δv / t trap.

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.
Keep these straight

a = (v − u) ÷ t.

01
[2 marks]

Define acceleration and state its SI unit.

  • Acceleration is the change in velocity per unit time. ✓
  • SI unit: metre per second squared (m/s²). ✓
02
Calculation
[2 marks]

A car speeds up from 8.0 m/s to 24 m/s in 4.0 s. Calculate its acceleration.

a = (v − u) ÷ t = (24 − 8.0) ÷ 4.0 = 16 ÷ 4.0

a = 4.0 m/s²

03
Calculation
[3 marks]

A cyclist slows from 20 m/s to rest in 5.0 s. Calculate the acceleration and explain what the sign of your answer means.

a = (0 − 20) ÷ 5.0 = −20 ÷ 5.0

a = −4.0 m/s²

  • The negative sign shows the cyclist is decelerating (slowing down). ✓
04
Analysis
[2 marks]

On a velocity-time graph, state what is represented by:

(a)the gradient of the line; (b)a line sloping downward.
  • (a) The gradient represents the acceleration. ✓
  • (b) A downward slope means the object is decelerating (negative acceleration). ✓
05
Calculation
[3 marks]

To find the acceleration of a car that goes from 5.0 m/s to 25 m/s in 4.0 s, a student divides 25 by 4.0. Explain the error and give the correct acceleration.

  • Acceleration uses the change in velocity, not the final velocity, so they must subtract u first. ✓
a = (25 − 5.0) ÷ 4.0 = 20 ÷ 4.0

a = 5.0 m/s² not 25 ÷ 4 = 6.25

06
Calculation
[2 marks]

A straight section of a velocity-time graph rises from 6.0 m/s at t = 0 to 26 m/s at t = 10 s. Calculate the acceleration over this section.

Gradient.

a = (26 − 6.0) ÷ (10 − 0) = 20 ÷ 10

a = 2.0 m/s²