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A-levelMechanics

Kinematics

Kinematics describes motion using displacement, velocity and acceleration. Choose a positive direction and use a graph or equation that fits the motion. SUVAT requires constant acceleration; signed area under a velocity–time graph gives displacement.

4 lessons6 worksheets18 questions

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Lessons

These free MathLab Learn guides are arranged in the order they are most useful for this stage.

  1. 01Position, displacement, velocity and acceleration
  2. 02Constant acceleration and SUVAT equations
  3. 03Kinematics graphs: displacement, velocity and acceleration
  4. 04Two dimensional motion with vectors

Practise

Practice moves from direct fluency towards application, reasoning and synthesis.

  1. Intro

    Position, Displacement And Average Rates

    Step 1 · 1 worksheet · 3 questions

  2. Intro

    Direct Constant-Acceleration Calculations

    Step 2 · 1 worksheet · 3 questions

  3. Practice

    Braking, Turning Points And Reversal

    Step 3 · 1 worksheet · 3 questions

  4. Practice

    Vertical Motion Under Gravity

    Step 4 · 1 worksheet · 3 questions

  5. Practice

    Velocity-Time Graphs

    Step 5 · 1 worksheet · 3 questions

  6. Challenge

    Constant Acceleration With Vectors

    Step 6 · 1 worksheet · 3 questions

By MathLab · Updated

Start with kinematics language if distance and displacement, or speed and velocity, are still easy to confuse.

Kinematics practice questions with answers

These three original MathLab examples move from a direct calculation to interpreting a reversal. Attempt them before opening the worked solutions.

Try a sample
01
MathLab practice

Foundation: read a velocity graph

Sample

A straight-line velocity–time graph runs from (0,2)(0,2) to (4,10)(4,10), with time in seconds and velocity in metres per second. Find the acceleration and displacement over these four seconds.

Worked answer: gradient and trapezium area

Acceleration is the gradient: (10−2)/4=2 m s−2(10-2)/4=\boxed{2\text{ m s}^{-2}}. Displacement is the area: 12(2+10)×4=24 m\tfrac12(2+10)\times4=\boxed{24\text{ m}}. Velocity stays positive, so distance is also 2424 m.

02
MathLab practice

Standard: choose a SUVAT equation

Sample

A cyclist moving at 3 m s−13\text{ m s}^{-1} accelerates uniformly at 2 m s−22\text{ m s}^{-2} over a displacement of 2020 m. Find the final speed.

Worked answer: use the equation without time

Use v2=u2+2asv^2=u^2+2as: v2=32+2(2)(20)=89v^2=3^2+2(2)(20)=89. The cyclist continues in the positive direction, so v=89 m s−1v=\sqrt{89}\text{ m s}^{-1}. The final speed is 9.43 m s−19.43\text{ m s}^{-1} to three significant figures.

03
MathLab practice

Reasoning: distance is not displacement

Sample

A particle has velocity v=4−2tv=4-2t for 0≤t≤40\le t\le4, in SI units. A student says it travels no distance because its final displacement is zero. Explain the error and find the distance.

Worked answer: split at the change of direction

Velocity changes sign at t=2t=2. The positive area is 12(2)(4)=4\tfrac12(2)(4)=4 m; the negative area is −4-4 m. They cancel for displacement, but distance adds their magnitudes: 4+4=84+4=8 m. The particle returns to its starting position after travelling away from it.

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