IGCSE Additional Mathematics: Kinematics Practice Questions
Differentiating displacement gives velocity, and differentiating velocity gives acceleration. Integrating reverses each step. A particle is instantaneously at rest when its velocity is zero.
Topic 17 of Cambridge IGCSE Additional Mathematics 0606 applies calculus to motion. The distinction between distance travelled and displacement is the question that separates candidates, because it requires checking for a change of direction. The questions below build to it.
What you need to know for Kinematics
- The calculus chainDisplacement differentiates to velocity, which differentiates to acceleration. Integration reverses the chain.
- At restA particle is instantaneously at rest when v = 0. Solve the velocity equation to find those times.
- Maximum or minimum velocityOccurs where the acceleration is zero, since acceleration is the derivative of velocity.
- Displacement from velocityIntegrate the velocity function. Use the initial condition, often s = 0 at t = 0, to find the constant.
- Distance against displacementIf the particle changes direction within the interval, split the integral at that time and add the sizes of the separate displacements.
IGCSE Additional Mathematics Kinematics questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
A particle moves in a straight line with velocity v = 3t2 minus 12t + 9 metres per second. Find the times at which the particle is instantaneously at rest.
Show the worked answer
- The particle is at rest when the velocity is zero.
- Set 3t2 minus 12t + 9 = 0 and divide through by 3: t2 minus 4t + 3 = 0.
- Factorise: (t minus 1)(t minus 3) = 0.
- So t = 1 s or t = 3 s. Both are positive, so both are valid times.
For the same particle, with v = 3t2 minus 12t + 9, find the acceleration when t = 1 and state what the sign of your answer means.
Show the worked answer
- Acceleration is the derivative of velocity with respect to time.
- a = dv over dt = 6t minus 12.
- At t = 1: a = 6 minus 12 = minus 6 m/s2.
- The negative sign means the acceleration acts in the negative direction. Since the particle is momentarily at rest at t = 1, it is about to move in the negative direction.
For the same particle, which starts at the origin, find an expression for the displacement s in terms of t.
Show the worked answer
- Displacement is the integral of velocity with respect to time.
- Integrate 3t2 minus 12t + 9 term by term: t3 minus 6t2 + 9t + c.
- The particle starts at the origin, so s = 0 when t = 0.
- Substituting gives 0 = 0 minus 0 + 0 + c, so c = 0 and s = t3 minus 6t2 + 9t.
For the same particle, find the total distance travelled in the first 2 seconds, and explain why it differs from the displacement.
Show the worked answer
- The particle is at rest at t = 1, so it changes direction there. The interval must be split at t = 1.
- From t = 0 to t = 1: s = 1 minus 6 + 9 = 4 m, so the particle moves 4 m in the positive direction.
- At t = 2: s = 8 minus 24 + 18 = 2 m. Between t = 1 and t = 2 the displacement changes from 4 to 2, a movement of 2 m in the negative direction.
- Total distance travelled = 4 + 2 = 6 m, adding the sizes of both movements.
- The displacement is only 2 m, because the return journey cancels part of the outward one. Distance ignores direction, displacement does not.
Common mistakes in this topic
- Confusing at rest with zero acceleration.
- Integrating when differentiation is required, or the reverse.
- Omitting the constant of integration or failing to evaluate it.
- Treating displacement as distance when the particle changes direction.
- Losing the units, which differ at each stage of the chain.
Exam tips
- Draw the chain before starting: displacement, velocity, acceleration, with differentiation one way and integration the other.
- At rest means v = 0. Maximum velocity means a = 0. Two different conditions.
- Always check for a change of direction before answering a distance question.
- Use the initial condition immediately after integrating, while the constant is still visible.
- Quote units at every stage: metres, metres per second, metres per second squared.
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Kinematics FAQs
What is the difference between distance and displacement?
Displacement is the net change in position, taking direction into account, and is found by evaluating the displacement function. Distance is the total path length travelled, ignoring direction, so if the particle reverses within the interval the journey must be split and the sizes added.
How do I know when a particle is at rest?
Set the velocity equal to zero and solve for time. A particle at rest has zero velocity, not zero acceleration. Zero acceleration instead indicates a maximum or minimum velocity, which is a different condition entirely.
How do I find displacement from a velocity function?
Integrate the velocity with respect to time, which gives the displacement plus a constant. Use the initial condition, usually that displacement is zero when time is zero, to evaluate that constant and complete the expression.
When does a particle have maximum velocity?
When its acceleration is zero, since acceleration is the derivative of velocity and a maximum or minimum occurs where the derivative vanishes. Differentiate the velocity, set the result to zero, and solve for the time.
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.