A-Level Kinematics graphs Practice Questions

Free A-Level Kinematics graphs practice questions with full step-by-step worked solutions. Covers v-t graph, area under graph, displacement, rectangle area. Practise exam-style problems and check your method.

v-t grapharea under graphdisplacementrectangle areaconstant velocitygradient
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A velocity-time graph for a particle is a single straight line rising from the origin (0,0)(0,0) to the point (4,12)(4,12), where time tt is in seconds and velocity vv is in m/s. Find the displacement of the particle during the first 44 seconds.
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Worked solution

  1. Identify the region under the velocity-time graph

    triangle: base 4 s, height 12 m/s\text{triangle: base } 4\text{ s, height } 12\text{ m/s}

    Displacement is the area between the line and the time-axis.

  2. Apply the area formula for a triangle

    s=12×4×12s = \tfrac12 \times 4 \times 12

    Area of a triangle is half the base times the height.

  3. Evaluate the displacement

    s=24 ms = 24\text{ m}

    This is the displacement over the interval.

Answer
24 m24\text{ m}
Question 2
2 markseasy
On a velocity-time graph a segment has a positive gradient while the velocity is positive. What is happening?
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Worked solution

  1. The gradient of a v-t graph is the acceleration

    a=gradient of v-ta = \text{gradient of } v\text{-}t

    A positive gradient means positive acceleration.

  2. Velocity and acceleration act in the same direction

    v>0, a>0v>0,\ a>0

    Both are positive.

  3. So the object is speeding up

    ∣v∣ increasing|v| \text{ increasing}

    The speed is growing over time.

Answer
The velocity is increasing, so the object is speeding up
Question 3
3 marksintermediate
A displacement-time graph is a straight line from (0,0)(0,0) to (4,20)(4,20), then a horizontal line from (4,20)(4,20) to (9,20)(9,20). Which description matches the motion?
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Worked solution

  1. Analyse the first segment of the d-t graph

    v1=20−04−0=5 m/sv_1 = \frac{20-0}{4-0} = 5\text{ m/s}

    A positive constant gradient means constant velocity.

  2. Analyse the second segment

    v2=20−209−4=0 m/sv_2 = \frac{20-20}{9-4} = 0\text{ m/s}

    A horizontal d-t line means at rest.

  3. Phase 1 is uniform motion away from the start

    v1=5 m/s (constant)v_1 = 5\text{ m/s (constant)}

    Straight sloped line = constant speed.

  4. Phase 2 is stationary

    v2=0v_2 = 0

    Displacement no longer changes.

  5. The object does not return towards the start

    s never decreasess \text{ never decreases}

    Displacement stays at 20 m.

  6. Combine the two phases

    move at 5 m/s, then stop\text{move at }5\text{ m/s, then stop}

    First constant velocity, then at rest.

Answer
Moves at a constant 5 m/s for 4 s, then remains stationary
Question 4
5 markshard
A displacement-time graph is a smooth curve that gets steeper as time increases (concave up). What does this shape tell you about the motion?
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Worked solution

  1. Velocity is the gradient of a d-t graph

    v=dsdtv = \frac{ds}{dt}

    Read the slope to find velocity.

  2. A steepening curve means the gradient is increasing

    gradient increasing\text{gradient increasing}

    Each later tangent is steeper than the last.

  3. Increasing gradient means increasing velocity

    v increasingv \text{ increasing}

    The object goes faster over time.

  4. Increasing velocity means positive acceleration

    a>0a > 0

    Acceleration is the rate of change of velocity.

  5. The curve is not straight, so velocity is not constant

    not linear\text{not linear}

    A straight d-t line would mean constant velocity.

  6. It is not horizontal, so the object is not at rest

    gradient≠0\text{gradient} \ne 0

    There is motion throughout.

  7. The gradient never decreases here

    no slowing shown\text{no slowing shown}

    So the object is not decelerating.

  8. Displacement keeps increasing

    s increasings \text{ increasing}

    Motion is away from the start.

  9. Combine the deductions

    v↑, a>0v\uparrow,\ a>0

    Speeding up with positive acceleration.

  10. Conclude the description

    accelerating\text{accelerating}

    The object is speeding up.

Answer
The object is accelerating (velocity increasing)
Question 5
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?
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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

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