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Constant acceleration (suvat) equationsEdexcel A-Level Maths: Subtopic test

10 questions, 27 marks

Edexcel A-Level Maths

Constant acceleration (suvat) equations

Total 27 marks

Name

Class

Date

  1. 1
    A ball is thrown vertically upwards from ground level with speed 1414 m s−1^{-1}. The ball is modelled as a particle moving freely under gravity, with g=9.8g=9.8 m s−2^{-2} and upwards taken as positive.
    (a)
    What is the greatest height reached by the ball?
    [1 mark]
    • A1010 m
    • B2020 m
    • C1.431.43 m
    • D9.19.1 m
    (b)
    How long is the ball in the air before it returns to ground level?
    [1 mark]
    • A0.710.71 s
    • B5.715.71 s
    • C2.862.86 s
    • D1.431.43 s
    (c)
    Find the speed of the ball 1.5 s after it is thrown, and state whether it is moving upwards or downwards.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A car brakes with constant deceleration from a speed of 3030 m s−1^{-1} and comes to rest after travelling 90 m in a straight line.
    (a)
    What is the magnitude of the car's deceleration?
    [1 mark]
    • A0.330.33 m s−2^{-2}
    • B55 m s−2^{-2}
    • C1010 m s−2^{-2}
    • D2.52.5 m s−2^{-2}
    (b)
    The car has travelled 40 m since it began to brake. What is its speed at that instant?
    [1 mark]
    • A500500 m s−1^{-1}
    • B36.136.1 m s−1^{-1}
    • C26.526.5 m s−1^{-1}
    • D22.422.4 m s−1^{-1}
    (c)
    Find the distance the car travels in the first 2 s of braking.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A particle moves in a straight line with constant acceleration aa m s−2^{-2}. At t=0t=0 its velocity is uu m s−1^{-1}. At time tt seconds its velocity is vv m s−1^{-1} and its displacement from its starting point is ss metres.
    (a)
    Using the fact that the area under a velocity--time graph is the displacement, show that s=ut+12at2s=ut+\frac12at^2.
    [3 marks]
    (b)
    In one case u=4u=4, a=2a=2 and s=21s=21. Find vv and tt.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A speeding car QQ passes a point AA at a constant speed of 2525 m s−1^{-1}. At that instant a police car PP, at rest at AA, starts to move in the same direction with constant acceleration 44 m s−2^{-2}. Time tt is measured in seconds from this instant.
    (a)
    (i) Write down expressions for the distances travelled by QQ and by PP after tt seconds.
    (ii) Find the time at which
    PP catches QQ.
    (iii) Find the speed of
    PP at that instant.
    [6 marks]
    (b)
    Suppose instead that PP cannot exceed 4040 m s−1^{-1}, and travels at this constant speed once it reaches it. Find the time at which PP catches QQ, and the distance from AA at which this happens.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).