A-level · Kinematics

Motion graphs (displacement–time & velocity–time)Exam-style questions

What the gradient and the area mean on each kind of motion graph — gradient of displacement–time is velocity, gradient of velocity–time is acceleration, area under velocity–time is displacement — and how to model a journey as a graph and read unknowns off its area.

12 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 3Level 4Level 5
  1. 1
    A straight horizontal road passes through the points AA and BB.825Vt (s)speed (m s⁻¹)Figure 1 (not accurately drawn)

    A van PP leaves AA from rest at time t=0t = 0 and travels along the road towards BB, where tt is measured in seconds. Figure 1 shows the speed-time graph for the motion of the van for 0⩽t⩽250 \leqslant t \leqslant 25
    • The van accelerates uniformly from rest for 88 seconds, reaching a speed of V m s−1V\ \text{m s}^{-1}
    • The van then moves at the constant speed V m s−1V\ \text{m s}^{-1}
    • The van passes BB when t=25t = 25
    The distance from AA to BB is 336336 m.
    (a) Show that V=16V = 16

    (3 marks)
    (b) Find the acceleration of the van during the first 88 seconds of its motion.

    (2 marks)

    After passing BB, the van continues along the road at the constant speed V m s−1V\ \text{m s}^{-1}

    A car QQ leaves AA from rest when t=5t = 5 and travels along the road in the same direction as the van. The car accelerates uniformly for TT seconds until its speed is 20 m s−120\ \text{m s}^{-1}, and then moves at this constant speed. The car draws level with the van when t=45t = 45

    The van and the car are modelled as particles.
    (c) On a copy of Figure 1, sketch speed-time graphs for the motion of the van and the motion of the car for 0⩽t⩽450 \leqslant t \leqslant 45

    (3 marks)
    (d) Find the value of TT

    (5 marks)
    Level 5 · 13 marksTry it