AS & A Level · Practice questions · Displacement, velocity and acceleration graphs

Read the graph.

Six original Cambridge-style questions on displacement-time and velocity-time graphs, 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

Sketch it, then gradients and areas.

Sketch the velocity-time graph, marking every time and velocity you know.
Gradient of \(v\)-\(t\) is acceleration; of \(s\)-\(t\) is velocity.
Area under \(v\)-\(t\) is displacement: triangles, rectangles, trapezia.
Sign: area below the axis is negative displacement but adds to distance.
Two particles: write each displacement in terms of \(t\) and set them equal.

Recall

[3 marks]

A runner completes one lap of a 400 m track in 80 s, finishing where she started. Find her distance travelled, displacement, average speed and average velocity.

Distance 400 m; displacement 0. ✓
Average speed \(= 400 \div 80 = 5\ \text{m s}^{-1}\). ✓
Average velocity \(= 0 \div 80 = 0\). ✓

Calculation

[4 marks]

A displacement-time graph consists of straight lines joining \((0, 0)\), \((5, 30)\), \((8, 30)\) and \((12, -10)\), with \(t\) in seconds and \(s\) in metres. Find the velocity in each stage, the total distance travelled and the final displacement.

Velocities: \(\dfrac{30}{5} = 6\), then 0, then \(\dfrac{-10 - 30}{4} = -10\ \text{m s}^{-1}\). ✓
Distance \(= 30 + 40 = 70\) m. ✓
Final displacement \(= -10\) m (10 m behind the start). ✓

Calculation

[4 marks]

A tram starts from rest and accelerates at \(1.5\ \text{m s}^{-2}\) for 10 s. It then travels at constant speed for 40 s before decelerating at \(2.5\ \text{m s}^{-2}\) to rest. Sketch the velocity-time graph and find the total distance.

Top speed \(= 1.5 \times 10 = 15\ \text{m s}^{-1}\); deceleration time \(= 15 \div 2.5 = 6\) s. ✓
Graph: up to 15 at \(t = 10\), flat to \(t = 50\), down to 0 at \(t = 56\). ✓
Distance \(= 75 + 600 + 45 = 720\) m. ✓

Analysis

[4 marks]

A car starts from rest and accelerates uniformly to a speed \(V\ \text{m s}^{-1}\) in 12 s. It travels at this speed for 30 s and then decelerates uniformly to rest in 8 s. The total distance is 1000 m. Find \(V\) and the acceleration.

Area \(= \tfrac{1}{2}(12)V + 30V + \tfrac{1}{2}(8)V = 40V\). ✓
\(40V = 1000\), so \(V = 25\). ✓
Acceleration \(= 25 \div 12 = 2.08\ \text{m s}^{-2}\). ✓

Analysis

[5 marks]

A particle moves in a straight line. Its velocity decreases uniformly from \(8\ \text{m s}^{-1}\) at \(t = 0\) to \(-6\ \text{m s}^{-1}\) at \(t = 7\). Find (a) the time at which it is instantaneously at rest, (b) the distance travelled in the 7 s, (c) its displacement at \(t = 7\).

Acceleration \(= \dfrac{-6 - 8}{7} = -2\ \text{m s}^{-2}\), so \(8 - 2t = 0\) at (a) \(t = 4\). ✓
Areas: \(\tfrac{1}{2}(4)(8) = 16\) m above, \(\tfrac{1}{2}(3)(6) = 9\) m below. ✓
(b) Distance \(= 16 + 9 = 25\) m. ✓
(c) Displacement \(= 16 - 9 = 7\) m. ✓

Analysis

[6 marks]

Car A starts from rest at a junction O and accelerates at \(0.8\ \text{m s}^{-2}\) for 20 s, then continues at constant speed. At the instant A starts, car B passes O travelling in the same direction at a constant \(12\ \text{m s}^{-1}\). Find the time when A overtakes B and the distance from O.

While accelerating, A's displacement is \(0.4t^2\). Setting \(0.4t^2 = 12t\) gives \(t = 30\), which is after the 20 s, so A has not caught B by then. ✓
At \(t = 20\), A has travelled \(0.4 \times 400 = 160\) m and has speed 16 \(\text{m s}^{-1}\). ✓
For \(t \ge 20\): \(160 + 16(t - 20) = 12t\), so \(4t = 160\), \(t = 40\) s. ✓
Distance \(= 12 \times 40 = 480\) m from O. ✓

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.