A-Level Constant acceleration Practice Questions

Free A-Level Constant acceleration practice questions with full step-by-step worked solutions. Covers suvat, kinematics, acceleration, equation-choice. Practise exam-style problems and check your method.

suvatkinematicsaccelerationequation-choicedisplacementvertical-motion
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A particle moves in a straight line with constant acceleration 2 m s−22\,\text{m s}^{-2}. Its initial velocity is 8 m s−18\,\text{m s}^{-1}. Find its velocity after 5 s5\,\text{s}.
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Worked solution

  1. List the known quantities and the unknown

    u=8 m s−1,  a=2 m s−2,  t=5 su=8\,\text{m s}^{-1},\; a=2\,\text{m s}^{-2},\; t=5\,\text{s}

    The initial velocity, acceleration and time are given; the final velocity is required.

  2. Select the equation linking u,a,tu,a,t and vv

    v=u+atv=u+at

    This equation connects exactly these four quantities.

  3. Substitute and evaluate

    v=8+2×5=18 m s−1v=8+2\times 5=18\,\text{m s}^{-1}

    Doing the arithmetic gives the final velocity.

Answer
v=18 m s−1v=18\,\text{m s}^{-1}
Question 2
2 markseasy
A velocity-time graph is a horizontal straight line lying above the time axis. What does this tell you about the motion?
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Worked solution

  1. Read the shape of the graph

    gradient=0\text{gradient}=0

    A horizontal line has zero slope.

  2. Interpret the gradient

    a=gradient=0a=\text{gradient}=0

    The gradient of a velocity-time graph is the acceleration.

  3. Interpret the height

    v=constant>0v=\text{constant}>0

    The velocity stays the same and is positive.

Answer
constant velocity\text{constant velocity}
Question 3
3 marksintermediate
Which constant-acceleration equation gives the displacement directly from the initial velocity uu, the acceleration aa and the time tt?
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Worked solution

  1. List what is known and unknown

    u,a,t known, v unknownu,a,t\ \text{known},\ v\ \text{unknown}

    We want displacement without the final velocity.

  2. Consider s=12(u+v)ts=\tfrac12(u+v)t

    needs v\text{needs } v

    This equation requires the final velocity, so reject it.

  3. Consider v2=u2+2asv^2=u^2+2as

    needs v too\text{needs } v\ \text{too}

    This also needs the final velocity.

  4. Consider v=u+atv=u+at

    no s\text{no } s

    This has no displacement at all.

  5. Consider s=ut+12at2s=ut+\tfrac12at^2

    uses u,a,t\text{uses } u,a,t

    This uses exactly the given quantities.

  6. Select the correct equation

    s=ut+12at2s=ut+\tfrac12 at^2

    It gives ss from u,a,tu,a,t.

Answer
s=ut+12at2s=ut+\tfrac12 at^2
Question 4
5 markshard
A ball is thrown vertically upwards at 29.4 m s−129.4\,\text{m s}^{-1}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find its height above the point of projection 2 s2\,\text{s} after launch.
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Worked solution

  1. Set up the sign convention

    up=+,  u=29.4 m s−1,  a=−9.8 m s−2\text{up}=+,\; u=29.4\,\text{m s}^{-1},\; a=-9.8\,\text{m s}^{-2}

    Upward positive.

  2. State what is wanted

    s at t=2 ss\ \text{at}\ t=2\,\text{s}

    Height after 2 seconds.

  3. Choose the displacement equation

    s=ut+12at2s=ut+\tfrac12 at^2

    Displacement from time.

  4. Substitute the values

    s=29.4×2−4.9×22s=29.4\times 2-4.9\times 2^2

    Insert u,a,tu,a,t.

  5. Compute each term

    s=58.8−19.6s=58.8-19.6

    The rise and the fall-back contributions.

  6. Subtract

    s=39.2 ms=39.2\,\text{m}

    The height above the start.

  7. Find the velocity there

    v=29.4−9.8×2=9.8v=29.4-9.8\times 2=9.8{}

    Velocity is positive (m/s).

  8. Interpret the velocity

    v>0⇒still risingv>0\Rightarrow\text{still rising}

    The ball has not yet reached the top.

  9. Confirm the units

    s in metress\ \text{in metres}

    The answer is a length.

  10. State the height

    s=39.2 ms=39.2\,\text{m}

    The ball is 39.2 m above the launch point.

Answer
s=39.2 ms=39.2\,\text{m}
Question 5
8 markschallenging
A velocity-time graph is made of three straight segments: a rise from 00 to 20 m s−120\,\text{m s}^{-1} over the first 10 s10\,\text{s}, a horizontal section at 20 m s−120\,\text{m s}^{-1} for the next 15 s15\,\text{s}, and a fall from 20 m s−120\,\text{m s}^{-1} to 00 over the final 10 s10\,\text{s}. Which one of the following statements about the motion is correct?
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Worked solution

  1. Split the graph into phases

    rise, level, fall\text{rise, level, fall}

    Three straight segments to analyse.

  2. Gradient of phase 1

    gradient=20−010\text{gradient}=\frac{20-0}{10}

    Rise over run for the first segment.

  3. Evaluate the phase 1 acceleration

    a1=2 m s−2a_1=2\,\text{m s}^{-2}

    Not 3 m/s^2.

  4. Gradient of phase 2

    gradient=0\text{gradient}=0

    Horizontal: constant velocity, not at rest.

  5. Interpret phase 2

    v=20 m s−1 constantv=20\,\text{m s}^{-1}\ \text{constant}

    Average speed is not 20 m/s over the whole journey, though.

  6. Gradient of phase 3

    0−2010=−2\frac{0-20}{10}=-2

    Deceleration is 2 m/s^2, not 5 m/s^2.

  7. Phase 1 area

    12×10×20=100\tfrac12\times 10\times 20=100

    Distance in phase 1 (metres).

  8. Phase 2 area

    20×15=30020\times 15=300

    Distance in phase 2 (metres).

  9. Phase 3 area

    12×10×20=100\tfrac12\times 10\times 20=100

    Distance in phase 3 (metres).

  10. Total distance

    100+300+100=500100+300+100=500

    Sum of the areas (metres).

  11. Reject the 600 m option

    500≠600500\neq 600

    The total distance is 500 m, not 600 m.

  12. Total time and average speed

    T=35, vˉ=50035=14.3T=35,\ \bar v=\frac{500}{35}=14.3

    Average speed is 14.3 m/s, not 20 m/s.

  13. Reject the at-rest option

    phase 2 is moving\text{phase 2 is moving}

    The object moves at constant speed in the middle.

  14. Reject the wrong deceleration

    decel=2≠5\text{decel}=2\neq 5

    Deceleration is 2 m/s^2.

  15. Select the correct statement

    s=500s=500{}

    The total distance travelled is 500 m.

Answer
s=500 ms=500\,\text{m}

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