AS & A Level · Practice questions · Constant acceleration: the suvat equations

Three known, two to find.

Six original Cambridge-style questions on motion with constant acceleration, including vertical motion with \(g = 10\), with full worked solutions.

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.

The method

List s, u, v, a, t, then choose.

Positive direction: choose it and give every vector a sign.
List \(s, u, v, a, t\): which three do you know, which one do you want?
Choose the equation that leaves out the fifth.
Gravity: \(a = -10\) with up positive, all the way up and down.
Stages: the end of one stage is the start of the next.

Recall

[3 marks]

A car travelling at \(8\ \text{m s}^{-1}\) accelerates uniformly at \(1.5\ \text{m s}^{-2}\) for 6 s. Find its final speed and the distance travelled.

\(v = 8 + 1.5 \times 6 = 17\ \text{m s}^{-1}\). ✓
\(s = 8 \times 6 + \tfrac{1}{2}(1.5)(36) = 48 + 27 = 75\) m. ✓

Calculation

[3 marks]

A train decelerates uniformly from \(30\ \text{m s}^{-1}\) to \(12\ \text{m s}^{-1}\) over a distance of 420 m. Find the deceleration and the time taken.

\(v^2 = u^2 + 2as\): \(144 = 900 + 840a\), so \(a = -0.9\). Deceleration \(0.9\ \text{m s}^{-2}\). ✓
\(s = \tfrac{1}{2}(u + v)t\): \(420 = 21t\), so \(t = 20\) s. ✓

Calculation

[5 marks]

A stone is projected vertically upwards at \(18\ \text{m s}^{-1}\) from ground level. Find its greatest height and the length of time for which it is more than 10 m above the ground.

Greatest height: \(0 = 324 - 20s\), so \(s = 16.2\) m. ✓
At 10 m: \(10 = 18t - 5t^2\), so \(5t^2 - 18t + 10 = 0\). ✓
\(t = \dfrac{18 \pm \sqrt{124}}{10}\), giving \(t = 0.686\) and \(t = 2.914\). ✓
Time above 10 m \(= \dfrac{2\sqrt{124}}{10} = 2.23\) s. ✓

Analysis

[5 marks]

A particle moves with constant acceleration. It travels 20 m in the 2nd second of its motion and 32 m in the 4th second. Find its initial velocity and its acceleration.

Distance in the \(n\)th second \(= s(n) - s(n-1) = u + a\left(n - \tfrac{1}{2}\right)\). ✓
2nd second: \(u + 1.5a = 20\). 4th second: \(u + 3.5a = 32\). ✓
Subtract: \(2a = 12\), so \(a = 6\ \text{m s}^{-2}\) and \(u = 20 - 9 = 11\ \text{m s}^{-1}\). ✓

Analysis

[5 marks]

A cyclist starts from rest and accelerates uniformly at \(0.5\ \text{m s}^{-2}\) until her speed is \(6\ \text{m s}^{-1}\). She then continues at this speed. Find the time she takes to travel 300 m from the start.

Stage 1: \(t = 6 \div 0.5 = 12\) s; \(s = \tfrac{1}{2}(0 + 6)(12) = 36\) m. ✓
Stage 2: remaining \(300 - 36 = 264\) m at \(6\ \text{m s}^{-1}\) takes 44 s. ✓
Total time \(= 12 + 44 = 56\) s. ✓

Analysis

[6 marks]

A ball is dropped from rest from a window 20 m above the ground. One second later a second ball is thrown vertically downwards from the same window with speed \(V\ \text{m s}^{-1}\). Both balls reach the ground at the same instant. Find \(V\).

First ball: \(20 = 5t^2\), so it lands at \(t = 2\) s. ✓
The second ball therefore takes \(2 - 1 = 1\) s. ✓
Down positive: \(20 = V(1) + \tfrac{1}{2}(10)(1)^2 = V + 5\). ✓
\(V = 15\). ✓

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.