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AS · Practice questions · Equations of motion

List, choose, solve.

Six original Cambridge-style questions on selecting and applying the suvat equations, on free-fall drops and throws, and on measuring g from a graph.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Constant acceleration only.

01
Analysis
[3 marks]

A cyclist accelerates uniformly from rest at 1.5 m s⁻² for 6.0 s. Find the final velocity and the distance travelled.

  • v = u + at = 0 + 1.5 × 6.0 = 9.0 m s⁻¹. ✓
  • s = ut + ½at² = 0 + ½ × 1.5 × 6.0² = 27 m. ✓
02
Analysis
[3 marks]

A train travelling at 40 m s⁻¹ brakes uniformly and stops in a distance of 500 m. Choosing a suitable equation, find the deceleration.

  • Known: u = 40, v = 0, s = 500; time not needed, so use v² = u² + 2as. ✓
  • 0 = 40² + 2a(500), so a = −1600 / 1000 = −1.6 m s⁻². ✓
  • The deceleration is 1.6 m s⁻². ✓
03
Analysis
[3 marks]

A ball is dropped from rest from a height of 1.8 m. Taking g = 9.8 m s⁻² and ignoring air resistance, find the time to reach the ground and the speed on landing.

  • s = ½gt²: 1.8 = ½ × 9.8 × t², so t² = 0.367, t = 0.61 s. ✓
  • v = gt = 9.8 × 0.61 = 6.0 m s⁻¹ (or v = √(2gs) = √(2 × 9.8 × 1.8) = 5.9 m s⁻¹). ✓
04
Analysis
[3 marks]

A ball is thrown vertically upward at 19.6 m s⁻¹. Taking g = 9.8 m s⁻², find the maximum height reached and the time taken to reach it.

  • At the top v = 0. Using v² = u² + 2as with a = −9.8: 0 = 19.6² − 2(9.8)s. ✓
  • s = 384 / 19.6 = 19.6 m. ✓
  • Time to top: v = u + at gives 0 = 19.6 − 9.8t, so t = 2.0 s. ✓
05
Analysis
[3 marks]

In an experiment to find g, a ball falls from rest through several measured heights, and a graph of fall distance s against time squared t² is plotted. The line is straight through the origin with gradient 4.85 m s⁻². Find g and explain why this graphical method is better than a single drop.

  • s = ½g t², so the gradient equals ½g; therefore g = 2 × 4.85 = 9.7 m s⁻². ✓
  • Using many points averages out random error and lets anomalies be spotted, giving a more reliable g than one reading. ✓
06
Analysis
[2 marks]

A car covers two stages: 12 s at a constant 20 m s⁻¹, then it accelerates uniformly. Explain why a single suvat equation cannot be applied to the whole journey, and state what you must do instead.

  • The suvat equations require a single constant acceleration, but the acceleration changes between the two stages (zero, then non-zero). ✓
  • Treat each stage separately with its own values, then combine the results (for example add the displacements). ✓

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.