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Constant acceleration practice problems Edexcel IAL Mechanics 1, Chapter 2

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 structured exam question

Displacement-time and velocity-time graphs, the five constant acceleration (suvat) formulae, and vertical motion under gravity. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Exam question practice
  5. Common mistakes
  6. Keep going

Key equations for constant acceleration

v = u + atno s
s = ½(u + v)tno a
v2 = u2 + 2asno t
s = ut + ½at2no v
s = vt − ½at2no u
area under a v-t graphdisplacement; the gradient is acceleration

Take g = 9.8 m s−2, as the WME01 paper asks, and give answers to 2 or 3 significant figures. Answers within 2.5% are marked correct, so rounding along the way is fine. Revise the methods in our A level Maths mechanics guide.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
A car speeds up uniformly from 4 m s−1 to 16 m s−1 in 6 s. Find its acceleration.
m s−2

Hint

You know u, v and t. Which suvat formula links them with a?

Worked solution
  1. Use v = u + at: 16 = 4 + 6a.
  2. 6a = 12, so a = 2 m s−2.
Q2Warm-up
A particle starts from rest and accelerates at 3 m s−2. How far does it travel in the first 5 s?
m

Hint

u = 0, a = 3, t = 5. You want s.

Worked solution
  1. Use s = ut + ½at2 with u = 0.
  2. s = ½ × 3 × 52 = 37.5 m.
Q3Warm-up
On a velocity-time graph, what does the area between the graph and the time axis represent?

Hint

Area is velocity multiplied by time.

Worked solution
  1. Velocity × time gives displacement.
  2. So the area is the displacement: B. The gradient is the acceleration.

Exam-style questions

Questions at the level of the Edexcel IAL Mechanics 1 exam. The ones marked Challenge combine more than one idea.

Q4Exam-styleVertical motion
A ball is projected vertically upwards at 14.7 m s−1. Modelling the ball as a particle moving freely under gravity, find the greatest height it reaches above the point of projection.
m

Hint

At the greatest height v = 0. Take upwards as positive, so a = −9.8.

Worked solution
  1. Upwards positive: u = 14.7, v = 0, a = −9.8.
  2. v2 = u2 + 2as: 0 = 14.72 − 2 × 9.8 × s.
  3. s = 216.09/19.6 = 11.0 m (3 s.f.).
Q5Exam-styleVertical motion
For the same ball, find the time from projection until it returns to the point of projection.
s

Hint

When it returns, the displacement s = 0.

Worked solution
  1. s = ut + ½at2: 0 = 14.7t − 4.9t2 = t(14.7 − 4.9t).
  2. t = 0 is the start, so t = 14.7/4.9 = 3 s.
Q6Exam-style
A cyclist moves with constant deceleration along a straight road. She passes A at 20 m s−1 and B at 8 m s−1, where AB = 84 m. Find the deceleration.
m s−2

Hint

You know u, v and s. Leave out t.

Worked solution
  1. v2 = u2 + 2as: 82 = 202 + 2 × a × 84.
  2. 64 = 400 + 168a, so a = −2.
  3. The deceleration is 2 m s−2.
Q7Exam-style
A stone is thrown vertically downwards at 4 m s−1 from the top of a cliff 50 m above the sea. Find its speed as it hits the sea.
m s−1

Hint

Take downwards as positive, so a = +9.8 and s = +50.

Worked solution
  1. Downwards positive: u = 4, a = 9.8, s = 50.
  2. v2 = 42 + 2 × 9.8 × 50 = 996.
  3. v = 31.6 m s−1 (3 s.f.).
Q8Challenge
Car P starts from rest at a set of traffic lights and accelerates at 2 m s−2. At the same moment car Q passes the lights at a constant 12 m s−1. How long is it until P catches Q?
s

Hint

When P catches Q they have both travelled the same distance from the lights.

Worked solution
  1. Distance for P: s = ½ × 2 × t2 = t2. Distance for Q: s = 12t.
  2. t2 = 12t, so t(t − 12) = 0.
  3. t = 0 is the start, so P catches Q after 12 s, 144 m from the lights.

Exam question practice

One original structured question in the style of the WME01 paper. Answer it in full, with working, before you open the mark scheme: M marks are for a correct method, A marks for accurate answers that follow from it, and B marks for a correct statement on its own.

Structured exam question style · 7 points · about 10 minutes

A train starts from rest at station A. It accelerates uniformly for 20 s to a speed of V m s−1, travels at this speed for 60 s, then decelerates uniformly for 10 s, coming to rest at station B. The distance from A to B is 1.8 km.

  1. (a) Sketch a speed-time graph for the journey.
  2. (b) Find the value of V.
  3. (c) Find the deceleration of the train.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

A trapezium: a straight line from (0, 0) up to (20, V), horizontal to (80, V), then straight down to (90, 0).

Part (b)

Area = ½(90 + 60)V = 75V = 1800, so V = 24.

Part (c)

Deceleration = 24/10 = 2.4 m s−2.

Your mark: 0 of 7

Common mistakes in constant acceleration

Keep going

Revise the ideas behind these questions:

Then try real exam questions: Edexcel IAL Mechanics 1 past papers, with official scoring guidelines. · All Edexcel IAL Mechanics 1 units