A-Level Maths kinematics: SUVAT, velocity-time graphs and variable acceleration

Where kinematics sits in each specification

Cambridge 9709 Paper 4 is 1 hour 15 minutes and 50 marks. Edexcel IAL M1 is 1 hour 30 minutes and 75 marks.

The key ideas

For constant acceleration: v = u + at, s = ut + (1/2)at^2, s = vt - (1/2)at^2, v^2 = u^2 + 2as, and s = (1/2)(u + v)t. Each equation leaves out one of the five quantities, so list what you know and what you want, and pick the equation that omits the quantity you neither know nor need.

Choose a positive direction at the start. For a ball thrown upwards, if up is positive then a = -g throughout the flight, including on the way down, and a displacement below the starting point is negative.

When acceleration is not constant, SUVAT does not apply. Use v = ds/dt and a = dv/dt, and integrate the other way: s is the integral of v, v is the integral of a, with a constant found from the initial conditions. Distance travelled is not always the same as displacement: if the particle changes direction, split the time interval where v = 0 and add the sizes of each part.

Worked example 1 (9709, g = 10): a ball is thrown upwards at 15 m s^-1 from 2 m above the ground

1. Take up as positive: u = 15, a = -10.
2. Greatest height above the point of projection: at the top v = 0, so 0 = 15^2 + 2(-10)s, which gives s = 225/20 = 11.25 m.
3. Time to hit the ground: the ground is 2 m below the start, so s = -2. Use s = ut + (1/2)at^2: -2 = 15t - 5t^2.
4. Rearrange: 5t^2 - 15t - 2 = 0, so t = (15 + sqrt(265))/10 = 3.13 s (taking the positive root).
5. With g = 9.8, as on an Edexcel paper that states it, the same method gives 11.5 m and 3.19 s. Always use the value the paper tells you.

Worked example 2 (9709 P4, Edexcel M2): v = 3t^2 - 12t + 9 for t ≥ 0. Find when the particle is at rest and the distance travelled in the first 3 seconds

1. At rest when v = 0: 3(t^2 - 4t + 3) = 3(t - 1)(t - 3) = 0, so t = 1 s and t = 3 s.
2. Displacement from the start: s = t^3 - 6t^2 + 9t (s = 0 at t = 0, so no constant).
3. At t = 1, s = 4; at t = 3, s = 27 - 54 + 27 = 0.
4. The particle moves 4 m forwards, stops, then moves 4 m back to the start.
5. So the displacement after 3 s is 0, but the distance travelled is 4 + 4 = 8 m.

Worked example 3: a velocity-time graph

1. A car accelerates uniformly from rest to 12 m s^-1 in 4 s, travels at 12 m s^-1 for 10 s, then decelerates uniformly to rest in 6 s.
2. Acceleration in the first stage: gradient = 12 ÷ 4 = 3 m s^-2. Deceleration in the last stage: 12 ÷ 6 = 2 m s^-2.
3. Distance = area under the graph = (1/2)(4)(12) + 10(12) + (1/2)(6)(12) = 24 + 120 + 36 = 180 m.

Common mistakes that cost marks

• Using SUVAT when the acceleration varies with time.
• Mixing signs: taking up as positive for velocity but using a = +10.
• Giving displacement when the question asks for distance, or the reverse.
• Rejecting the wrong root of a quadratic for time, or keeping a negative time.
• Using g = 9.8 on a Cambridge paper, or g = 10 on an Edexcel paper that states 9.8. The answer will not match the mark scheme.
• Reading the area under a speed-time graph as acceleration.

Exam technique and how a tutor helps

Start every kinematics question with a short diagram and a list: s = ?, u = 15, v = 0, a = -10, t = ?. It takes ten seconds, earns credit for a correct method, and makes the equation choice obvious. State the direction you took as positive. Give final answers to 3 significant figures unless the question says otherwise, and remember that answers on 9709 Paper 4 often come out exact because g = 10.

In one-to-one lessons a tutor focuses on setting up the problem, because kinematics marks are usually lost before any equation is written. Students practise translating words into a list of knowns, then multi-stage questions with two particles or a change of acceleration, using real past-paper questions from the board they sit.

Self-check: can you do these?

• A stone is dropped from rest and falls for 3 s (g = 10). How far does it fall? (Answer: 45 m)
• A car slows from 20 m s^-1 to 8 m s^-1 over 70 m. Find the deceleration. (Answer: 2.4 m s^-2)
• s = t^3 - 3t. Find the acceleration at t = 2. (Answer: 12)
• What does the gradient of a displacement-time graph represent? (Answer: velocity)
• Name the SUVAT equation without t. (Answer: v^2 = u^2 + 2as)