Hard A-Level Constant acceleration Questions

Challenging, exam-style A-Level Constant acceleration questions with worked solutions. Stretch yourself on the hardest multi-stage, suvat, vertical-motion, quadratic problems.

multi-stagesuvatvertical-motionquadraticspeedtwo-objects
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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}
Question 2
8 markschallenging
A vehicle accelerates from rest at 3 m s−23\,\text{m s}^{-2} for 8 s8\,\text{s}, then travels at constant speed for 30 s30\,\text{s}, then decelerates uniformly at 4 m s−24\,\text{m s}^{-2} to rest. Find the total distance travelled.
Show worked solution

Worked solution

  1. Model the motion in three phases

    accelerate, cruise, decelerate\text{accelerate, cruise, decelerate}

    Constant acceleration in each phase.

  2. Phase 1 knowns

    u=0 m s−1,  a=3 m s−2,  t=8 su=0\,\text{m s}^{-1},\; a=3\,\text{m s}^{-2},\; t=8\,\text{s}

    Acceleration from rest.

  3. Phase 1 cruising speed

    v=0+3×8=24v=0+3\times 8=24{}

    Speed reached (m/s).

  4. Phase 1 distance

    s1=12(0+24)×8=96 ms_1=\tfrac12(0+24)\times 8=96\,\text{m}

    Distance while accelerating.

  5. Phase 2 is constant speed

    v=24 m s−1,  t=30 sv=24\,\text{m s}^{-1},\; t=30\,\text{s}

    The cruising phase.

  6. Phase 2 distance

    s2=24×30=720 ms_2=24\times 30=720\,\text{m}

    Distance = speed times time.

  7. Phase 3 knowns

    u=24 m s−1,  v=0 m s−1,  a=−4 m s−2u=24\,\text{m s}^{-1},\; v=0\,\text{m s}^{-1},\; a=-4\,\text{m s}^{-2}

    The braking phase.

  8. Phase 3 time

    t3=0−24−4=6t_3=\frac{0-24}{-4}=6{}

    Time to stop (seconds).

  9. Phase 3 distance

    s3=12(24+0)×6=72 ms_3=\tfrac12(24+0)\times 6=72\,\text{m}

    Distance while braking.

  10. Collect the distances

    s=96+720+72s=96+720+72

    Add the phase distances.

  11. Evaluate the total distance

    s=888 ms=888\,\text{m}

    The total distance.

  12. Find the total time

    T=8+30+6=44T=8+30+6=44

    Total journey time (seconds).

  13. Confirm the units

    s in metress\ \text{in metres}

    The answer is a length.

  14. Sense-check

    888 m over 44 s888{}\ \text{m over }44{}\ \text{s}

    An average of about 20 m/s, which is reasonable.

  15. State the total distance

    s=888 ms=888\,\text{m}

    The vehicle travels 888 metres.

Answer
s=888 ms=888\,\text{m}
Question 3
8 markschallenging
A car travels at 20 m s−120\,\text{m s}^{-1}. The driver has a reaction time of 0.7 s0.7\,\text{s} before braking, during which the speed is unchanged. The brakes then decelerate the car uniformly at 5 m s−25\,\text{m s}^{-2} until it stops. Find the total stopping distance (thinking distance plus braking distance).
Show worked solution

Worked solution

  1. Split the stopping distance

    d=dthink+dbraked=d_{\text{think}}+d_{\text{brake}}

    Total = thinking distance + braking distance.

  2. During reaction the speed is constant

    v=20 m s−1v=20\,\text{m s}^{-1}

    No braking yet, so speed is unchanged.

  3. Thinking distance

    dthink=20×0.7d_{\text{think}}=20\times 0.7

    Distance = speed times reaction time.

  4. Evaluate the thinking distance

    dthink=14 md_{\text{think}}=14\,\text{m}

    The car travels 14 m before braking.

  5. Braking phase knowns

    u=20 m s−1,  v=0 m s−1,  a=−5 m s−2u=20\,\text{m s}^{-1},\; v=0\,\text{m s}^{-1},\; a=-5\,\text{m s}^{-2}

    Braking to rest.

  6. Choose v2=u2+2asv^2=u^2+2as

    v2=u2+2asv^2=u^2+2as

    Find the braking distance.

  7. Substitute the values

    0=202+2(−5)s0=20^2+2(-5)s

    Insert the numbers.

  8. Simplify

    0=400−10s0=400-10s

    Compute the terms.

  9. Solve for the braking distance

    s=40 ms=40\,\text{m}

    The braking distance.

  10. Find the braking time as a check

    t=205=4t=\frac{20}{5}=4

    Time to stop (seconds).

  11. Cross-check the braking distance

    12×20×4=40\tfrac12\times 20\times 4=40

    Distance agrees.

  12. Add the two distances

    d=14+40d=14+40

    Combine thinking and braking distances.

  13. Evaluate the total

    d=54 md=54\,\text{m}

    The total stopping distance.

  14. Confirm the units

    d in metresd\ \text{in metres}

    The answer is a length.

  15. State the total stopping distance

    d=54 md=54\,\text{m}

    The car needs 54 m to stop.

Answer
d=54 md=54\,\text{m}
Question 4
8 markschallenging
A ball is projected vertically upwards from ground level at 24.5 m s−124.5\,\text{m s}^{-1}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the total length of time for which the ball is at least 19.6 m19.6\,\text{m} above the ground.
Show worked solution

Worked solution

  1. Set up the sign convention

    up=+,  u=24.5 m s−1,  a=−9.8 m s−2\text{up}=+,\; u=24.5\,\text{m s}^{-1},\; a=-9.8\,\text{m s}^{-2}

    Upward positive.

  2. Find the times at height 19.6 m

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

    Use the displacement equation.

  3. Substitute s=19.6s=19.6

    19.6=24.5t−4.9t219.6=24.5t-4.9t^2

    Set the height to 19.6 m.

  4. Rearrange into standard form

    4.9t2−24.5t+19.6=04.9t^2-24.5t+19.6=0

    Bring all terms to one side.

  5. Divide through by 4.9

    t2−5t+4=0t^2-5t+4=0

    Simplify the quadratic.

  6. Factorise

    (t−1)(t−4)=0(t-1)(t-4)=0

    Factor the quadratic.

  7. Read off the roots

    t=1 s and t=4 st=1\,\text{s}\ \text{and}\ t=4\,\text{s}

    Two crossing times.

  8. Interpret the first crossing

    t=1 rising through 19.6t=1\ \text{rising through }19.6

    On the way up.

  9. Interpret the second crossing

    t=4 falling through 19.6t=4\ \text{falling through }19.6

    On the way down.

  10. Note the motion between them

    1<t<4⇒s>19.61<t<4\Rightarrow s>19.6

    Between these times it is above 19.6 m.

  11. Length of time above the height

    Δt=4−1\Delta t=4-1

    Difference of the crossing times.

  12. Evaluate the interval

    Δt=3 s\Delta t=3\,\text{s}

    It is above 19.6 m for 3 s.

  13. Check with the peak height

    smax⁡=24.5219.6=30.625s_{\max}=\frac{24.5^2}{19.6}=30.625

    Peak 30.6 m exceeds 19.6 m.

  14. Note the symmetry

    midpoint t=2.5\text{midpoint }t=2.5

    The two times are symmetric about the peak time.

  15. State the length of time

    Δt=3 s\Delta t=3\,\text{s}

    The ball is at least 19.6 m up for 3 s.

Answer
Δt=3 s\Delta t=3\,\text{s}
Question 5
8 markschallenging
A car starts from rest and accelerates uniformly at 2 m s−22\,\text{m s}^{-2} for 10 s10\,\text{s}. It then travels at constant speed for a further 200 m200\,\text{m}. Find the average speed for the whole journey.
Show worked solution

Worked solution

  1. Model the motion in two phases

    accelerate then constant speed\text{accelerate then constant speed}

    Two phases of motion.

  2. Phase 1 knowns

    u=0 m s−1,  a=2 m s−2,  t=10 su=0\,\text{m s}^{-1},\; a=2\,\text{m s}^{-2},\; t=10\,\text{s}

    Acceleration from rest.

  3. Phase 1 final speed

    v=0+2×10=20v=0+2\times 10=20{}

    Speed reached (m/s).

  4. Phase 1 distance

    s1=12(0+20)×10=100 ms_1=\tfrac12(0+20)\times 10=100\,\text{m}

    Distance while accelerating.

  5. Phase 2 is constant speed

    v=20 m s−1,  s2=200 mv=20\,\text{m s}^{-1},\; s_2=200\,\text{m}

    It cruises 200 m at 20 m/s.

  6. Phase 2 time

    t2=20020t_2=\frac{200}{20}

    Time = distance / speed.

  7. Evaluate phase 2 time

    t2=10 st_2=10\,\text{s}

    The cruise lasts 10 s.

  8. Total distance

    s=100+200s=100+200

    Add the two distances.

  9. Evaluate the total distance

    s=300 ms=300\,\text{m}

    Total distance travelled.

  10. Total time

    T=10+10T=10+10

    Add the two times.

  11. Evaluate the total time

    T=20 sT=20\,\text{s}

    Total journey time.

  12. Average-speed formula

    vˉ=total distancetotal time\bar v=\frac{\text{total distance}}{\text{total time}}

    Definition of average speed.

  13. Substitute the totals

    vˉ=30020\bar v=\frac{300}{20}

    Insert the totals.

  14. Evaluate the average speed

    vˉ=15 m s−1\bar v=15\,\text{m s}^{-1}

    The average speed.

  15. State the average speed

    average speed=15 m s−1\text{average speed}=15\,\text{m s}^{-1}

    The average speed for the journey is 15 m/s.

Answer
vˉ=15 m s−1\bar v=15\,\text{m s}^{-1}

Unlock 29 more Constant acceleration questions

Create a free account to work through every A-Level Constant acceleration question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Constant acceleration practice

Practice QuestionsWorked Solutions

Related Mechanics topics

Applications of forcesConnected particlesForces and frictionForces and Newton’s lawsFurther kinematicsKinematics graphs