Constant acceleration Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Constant acceleration questions. See exactly how to solve problems on suvat, kinematics, acceleration, equation-choice.

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}.

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 car starts from rest and accelerates uniformly at 3 m s−23\,\text{m s}^{-2} for 4 s4\,\text{s}. Find its final speed.

Worked solution

  1. Write down what is known

    u=0 m s−1,  a=3 m s−2,  t=4 su=0\,\text{m s}^{-1},\; a=3\,\text{m s}^{-2},\; t=4\,\text{s}

    Starting from rest means the initial velocity is zero.

  2. Choose the appropriate equation

    v=u+atv=u+at

    Velocity from initial velocity, acceleration and time.

  3. Substitute the values

    v=0+3×4=12 m s−1v=0+3\times 4=12\,\text{m s}^{-1}

    The final speed follows directly.

Answer
v=12 m s−1v=12\,\text{m s}^{-1}
Question 3
2 markseasy
A motorcycle increases its speed uniformly from 12 m s−112\,\text{m s}^{-1} to 30 m s−130\,\text{m s}^{-1} in 6 s6\,\text{s}. Find its acceleration.

Worked solution

  1. Identify the known quantities

    u=12 m s−1,  v=30 m s−1,  t=6 su=12\,\text{m s}^{-1},\; v=30\,\text{m s}^{-1},\; t=6\,\text{s}

    Two velocities and the time are known; acceleration is required.

  2. Rearrange v=u+atv=u+at for aa

    a=v−uta=\frac{v-u}{t}

    Make the acceleration the subject.

  3. Substitute and evaluate

    a=30−126=3 m s−2a=\frac{30-12}{6}=3\,\text{m s}^{-2}

    The acceleration is positive because the motorcycle speeds up.

Answer
a=3 m s−2a=3\,\text{m s}^{-2}
Question 4
2 markseasy
Which of the five constant-acceleration equations gives the displacement ss directly from uu, vv and tt when the acceleration is not known?

Worked solution

  1. State which quantities are available

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

    Only three quantities are given and aa is missing.

  2. Recall the standard equations

    v=u+at,  s=12(u+v)t,  v2=u2+2asv=u+at,\; s=\tfrac12(u+v)t,\; v^2=u^2+2as

    Compare which quantities each equation uses.

  3. Select the one that avoids aa

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

    Only this equation uses u,v,tu,v,t and not the acceleration.

Answer
s=12(u+v)ts=\tfrac12(u+v)t
Question 5
2 markseasy
A train accelerates uniformly from 10 m s−110\,\text{m s}^{-1} to 22 m s−122\,\text{m s}^{-1} over a period of 4 s4\,\text{s}. Find the distance travelled.

Worked solution

  1. List the knowns

    u=10 m s−1,  v=22 m s−1,  t=4 su=10\,\text{m s}^{-1},\; v=22\,\text{m s}^{-1},\; t=4\,\text{s}

    Both velocities and the time are given.

  2. Use the average-velocity equation

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

    This finds displacement from the two velocities and the time.

  3. Substitute and evaluate

    s=12(10+22)×4=64 ms=\tfrac12(10+22)\times 4=64\,\text{m}

    The train travels 64 metres.

Answer
s=64 ms=64\,\text{m}

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