Hard A-Level Kinematics graphs Questions

Challenging, exam-style A-Level Kinematics graphs questions with worked solutions. Stretch yourself on the hardest v-t graph, three-stage journey, average speed, distance problems.

v-t graphthree-stage journeyaverage speeddistancecomposite aread-t graph
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A particle's displacement-time graph has three straight sections: from (0,0)(0,0) to (4,16)(4,16); a horizontal section from (4,16)(4,16) to (7,16)(7,16); and from (7,16)(7,16) to (11,0)(11,0). Which velocity-time description matches this motion?
Show worked solution

Worked solution

  1. Velocity is the gradient of each d-t section

    v=ΔsΔtv = \frac{\Delta s}{\Delta t}

    Convert each section to a velocity.

  2. Section 1 gradient

    v1=16−04−0=4 m/sv_1 = \frac{16-0}{4-0} = 4\text{ m/s}

    Constant positive velocity.

  3. Section 2 gradient

    v2=16−167−4=0 m/sv_2 = \frac{16-16}{7-4} = 0\text{ m/s}

    At rest.

  4. Section 3 gradient

    v3=0−1611−7=−4 m/sv_3 = \frac{0-16}{11-7} = -4\text{ m/s}

    Constant negative velocity.

  5. Each section is straight, so each velocity is constant

    piecewise constant v\text{piecewise constant } v

    No curving means no acceleration within a section.

  6. The v-t graph is therefore a step-like set of horizontal lines

    v=4, 0, −4v = 4,\ 0,\ -4

    Three constant levels.

  7. Section 1 on the v-t graph

    v=+4 for 0≤t≤4v = +4\text{ for }0\le t\le 4

    A horizontal line at +4.

  8. Section 2 on the v-t graph

    v=0 for 4≤t≤7v = 0\text{ for }4\le t\le 7

    A horizontal line on the axis.

  9. Section 3 on the v-t graph

    v=−4 for 7≤t≤11v = -4\text{ for }7\le t\le 11

    A horizontal line at -4.

  10. The speed is the same going out and back

    ∣v1∣=∣v3∣=4|v_1| = |v_3| = 4

    Only the direction differs.

  11. It is not a single constant velocity

    v changes valuev \text{ changes value}

    So a single-level graph is wrong.

  12. It is not a smoothly sloping v-t line

    no uniform acceleration\text{no uniform acceleration}

    Gradients are piecewise constant, not steadily changing.

  13. The middle section is genuinely zero, not just small

    v2=0v_2 = 0

    The particle is stationary between t=4 and t=7.

  14. The final section is negative, not positive

    v3=−4<0v_3 = -4 < 0

    Motion reverses direction.

  15. Conclude the matching v-t description

    +4, 0, −4+4,\ 0,\ -4

    Constant +4, then 0, then constant -4.

Answer
+4+4 m/s for 4 s, then 00 m/s for 3 s, then −4-4 m/s for 4 s
Question 2
8 markschallenging
A cyclist's displacement-time graph goes straight from (0,0)(0,0) to (10,50)(10,50), stays at 5050 m to t=14t=14, then returns straight to (24,0)(24,0) (displacement in metres). What is the cyclist's average velocity over the whole 24 s?
Show worked solution

Worked solution

  1. Average velocity uses net displacement, not distance

    vˉvel=ΔsΔt\bar v_{vel} = \frac{\Delta s}{\Delta t}

    Net displacement is final minus initial position.

  2. Outbound velocity

    v1=5010=5 m/sv_1 = \frac{50}{10} = 5\text{ m/s}

    Positive gradient away from start.

  3. Rest-phase velocity

    v2=0v_2 = 0

    Horizontal segment.

  4. Return velocity

    v3=0−5024−14=−5 m/sv_3 = \frac{0-50}{24-14} = -5\text{ m/s}

    Negative gradient back to start.

  5. Outbound distance

    d1=50 md_1 = 50\text{ m}

    Distance covered going out.

  6. Return distance

    d3=50 md_3 = 50\text{ m}

    Distance covered coming back.

  7. Total distance travelled

    d=50+50=100 md = 50 + 50 = 100\text{ m}

    Used for average speed.

  8. Average speed for contrast

    vˉsp=10024=256 m/s\bar v_{sp} = \frac{100}{24} = \tfrac{25}{6}\text{ m/s}

    About 4.17 m/s.

  9. Net displacement over the journey

    Δs=0−0=0 m\Delta s = 0 - 0 = 0\text{ m}

    The cyclist returns to the starting point.

  10. Total time

    Δt=24 s\Delta t = 24\text{ s}

    Whole journey duration.

  11. Average velocity

    vˉvel=024=0 m/s\bar v_{vel} = \frac{0}{24} = 0\text{ m/s}

    Zero net displacement gives zero average velocity.

  12. Average velocity is not the average speed

    0≠2560 \ne \tfrac{25}{6}

    The two differ because of the return.

  13. It is not simply the outbound velocity

    0≠50 \ne 5

    Direction matters for velocity.

  14. It is not negative overall

    vˉvel=0\bar v_{vel} = 0

    Equal out and back cancel exactly.

  15. Conclude the average velocity

    vˉvel=0 m/s\bar v_{vel} = 0\text{ m/s}

    Because start and end positions coincide.

Answer
00 m/s, because the cyclist returns to the starting point
Question 3
8 markschallenging
A velocity-time graph is piecewise linear through (0,0)(0,0), (4,8)(4,8), (8,8)(8,8), (10,20)(10,20), (14,20)(14,20) and (18,0)(18,0) (time in s, velocity in m/s). Which single statement is correct?
Show worked solution

Worked solution

  1. Plan: find every gradient and area from the graph

    a=gradient, s=areaa=\text{gradient},\ s=\text{area}

    Work through each segment.

  2. Segment A gradient (0,0)→(4,8)(0,0)\to(4,8)

    aA=84=2 m/s2a_A = \frac{8}{4} = 2\text{ m/s}^2

    Initial acceleration.

  3. Segment B gradient (4,8)→(8,8)(4,8)\to(8,8)

    aB=0a_B = 0

    Constant velocity.

  4. Segment C gradient (8,8)→(10,20)(8,8)\to(10,20)

    aC=20−810−8=6 m/s2a_C = \frac{20-8}{10-8} = 6\text{ m/s}^2

    Steepest acceleration.

  5. Segment D gradient (10,20)→(14,20)(10,20)\to(14,20)

    aD=0a_D = 0

    Constant velocity again.

  6. Segment E gradient (14,20)→(18,0)(14,20)\to(18,0)

    aE=0−204=−5 m/s2a_E = \frac{0-20}{4} = -5\text{ m/s}^2

    Deceleration to rest.

  7. Greatest acceleration magnitude

    max⁡(2,0,6,0,5)=6\max(2,0,6,0,5) = 6

    This occurs on segment C, t=8 to t=10.

  8. Segment A area (triangle)

    AA=12×4×8=16 mA_A = \tfrac12 \times 4 \times 8 = 16\text{ m}

    First region.

  9. Segment B area (rectangle)

    AB=8×4=32 mA_B = 8 \times 4 = 32\text{ m}

    Second region.

  10. Segment C area (trapezium)

    AC=12(8+20)×2=28 mA_C = \tfrac12 (8+20)\times 2 = 28\text{ m}

    Third region.

  11. Segment D area (rectangle)

    AD=20×4=80 mA_D = 20 \times 4 = 80\text{ m}

    Fourth region.

  12. Segment E area (triangle)

    AE=12×4×20=40 mA_E = \tfrac12 \times 4 \times 20 = 40\text{ m}

    Fifth region.

  13. Total distance

    s=16+32+28+80+40=196 ms = 16+32+28+80+40 = 196\text{ m}

    So a claim of 100 m would be false.

  14. The velocity is not constant throughout

    aC≠0a_C \ne 0

    So a constant-velocity claim is false.

  15. Conclude the correct statement

    amax⁡ on (8,10)a_{\max} \text{ on } (8,10)

    Greatest acceleration is between t=8 and t=10.

Answer
The greatest acceleration occurs between t=8t=8 and t=10t=10
Question 4
8 markschallenging
A vehicle accelerates uniformly from rest to 2525 m/s in 44 s, cruises at 2525 m/s for an unknown time, then decelerates uniformly to rest in 44 s. On the velocity-time graph this is a triangle, a rectangle and a triangle. The total distance travelled is 400400 m. Find the length of time the vehicle cruises at constant velocity.
Show worked solution

Worked solution

  1. Describe the graph, marking the unknown cruise duration

    cruise time=tc (unknown)\text{cruise time} = t_c\ (\text{unknown})

    Three phases with the middle duration unknown.

  2. Stage 1: acceleration is the gradient

    a1=254=6.25 m/s2a_1 = \frac{25}{4} = 6.25\text{ m/s}^2

    Rise over run for the first line.

  3. Stage 1: displacement is a triangle area

    s1=12×4×25s_1 = \tfrac12 \times 4 \times 25

    Area under the acceleration segment.

  4. Stage 1: evaluate the displacement

    s1=50 ms_1 = 50\text{ m}

    Known distance for the first phase.

  5. Stage 3: deceleration is the gradient

    a3=0−254=−6.25 m/s2a_3 = \frac{0-25}{4} = -6.25\text{ m/s}^2

    Negative gradient for slowing down.

  6. Stage 3: displacement is a triangle area

    s3=12×4×25=50 ms_3 = \tfrac12 \times 4 \times 25 = 50\text{ m}

    Known distance for the last phase.

  7. Stage 2: cruise displacement in terms of tct_c

    s2=25 tcs_2 = 25\,t_c

    Rectangle area with unknown width.

  8. Write the total-distance equation

    s1+s2+s3=400s_1 + s_2 + s_3 = 400

    The three areas sum to the total distance.

  9. Substitute the known displacements

    50+25 tc+50=40050 + 25\,t_c + 50 = 400

    Insert the two triangle areas.

  10. Combine the constant displacement terms

    25 tc+100=40025\,t_c + 100 = 400

    Add the two known triangle areas.

  11. Isolate the cruise term

    25 tc=400−100=30025\,t_c = 400 - 100 = 300

    Subtract the known distance from the total.

  12. Solve for the cruise time

    tc=30025=12 st_c = \frac{300}{25} = 12\text{ s}

    Divide by the cruising velocity.

  13. State the cruise duration

    tc=12 st_c = 12\text{ s}

    This is the required constant-velocity time.

  14. Find the total journey time

    t=4+12+4=20 st = 4 + 12 + 4 = 20\text{ s}

    Sum of all three phase durations.

  15. Compute the average speed as a check

    vˉ=40020=20 m/s\bar v = \frac{400}{20} = 20\text{ m/s}

    Confirms the journey is consistent.

Answer
12 s12\text{ s}
Question 5
8 markschallenging
A vehicle accelerates uniformly from rest to 1515 m/s in 44 s, cruises at 1515 m/s for an unknown time, then decelerates uniformly to rest in 44 s. On the velocity-time graph this is a triangle, a rectangle and a triangle. The total distance travelled is 240240 m. Find the length of time the vehicle cruises at constant velocity.
Show worked solution

Worked solution

  1. Describe the graph, marking the unknown cruise duration

    cruise time=tc (unknown)\text{cruise time} = t_c\ (\text{unknown})

    Three phases with the middle duration unknown.

  2. Stage 1: acceleration is the gradient

    a1=154=3.75 m/s2a_1 = \frac{15}{4} = 3.75\text{ m/s}^2

    Rise over run for the first line.

  3. Stage 1: displacement is a triangle area

    s1=12×4×15s_1 = \tfrac12 \times 4 \times 15

    Area under the acceleration segment.

  4. Stage 1: evaluate the displacement

    s1=30 ms_1 = 30\text{ m}

    Known distance for the first phase.

  5. Stage 3: deceleration is the gradient

    a3=0−154=−3.75 m/s2a_3 = \frac{0-15}{4} = -3.75\text{ m/s}^2

    Negative gradient for slowing down.

  6. Stage 3: displacement is a triangle area

    s3=12×4×15=30 ms_3 = \tfrac12 \times 4 \times 15 = 30\text{ m}

    Known distance for the last phase.

  7. Stage 2: cruise displacement in terms of tct_c

    s2=15 tcs_2 = 15\,t_c

    Rectangle area with unknown width.

  8. Write the total-distance equation

    s1+s2+s3=240s_1 + s_2 + s_3 = 240

    The three areas sum to the total distance.

  9. Substitute the known displacements

    30+15 tc+30=24030 + 15\,t_c + 30 = 240

    Insert the two triangle areas.

  10. Combine the constant displacement terms

    15 tc+60=24015\,t_c + 60 = 240

    Add the two known triangle areas.

  11. Isolate the cruise term

    15 tc=240−60=18015\,t_c = 240 - 60 = 180

    Subtract the known distance from the total.

  12. Solve for the cruise time

    tc=18015=12 st_c = \frac{180}{15} = 12\text{ s}

    Divide by the cruising velocity.

  13. State the cruise duration

    tc=12 st_c = 12\text{ s}

    This is the required constant-velocity time.

  14. Find the total journey time

    t=4+12+4=20 st = 4 + 12 + 4 = 20\text{ s}

    Sum of all three phase durations.

  15. Compute the average speed as a check

    vˉ=24020=12 m/s\bar v = \frac{240}{20} = 12\text{ m/s}

    Confirms the journey is consistent.

Answer
12 s12\text{ s}

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