Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 4: Mechanics · official syllabus
KINEMATICS LAB
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Four short missions on signed area, acceleration, stopping and displacement-time curves. Predict the rule, then calculate and test your answer. These models describe motion in a straight line.
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Mission 1 of 4
Read signed area
A particle moves at +4 m/s for 3 s, then at -2 m/s for 2 s. Treat the change at t = 3 as instantaneous in this idealised model. The fixed velocity-time graph shows both stages. Set its signed displacement over the full 5 s, not its total distance travelled.
An area below the time axis on a velocity-time graph contributes...
White: the given velocity-time graph. Shading shows the two signed areas; the dashed jump is idealised.
Your displacement is 3.0 m. Subtract the magnitude of the below-axis area.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Choose a direction and distinguish scalars 4.2
Motion is in one dimension. Choose a positive direction and a fixed origin before writing signed displacements, velocities and accelerations. Distance and speed are scalar quantities; displacement, velocity and acceleration have direction.
Displacement is the change in position from start to finish. Distance is the length of the whole path. A journey 12 m forward then 4 m back has displacement +8 m but distance 16 m.
Speed is the magnitude of velocity. Negative velocity means motion in the negative direction, not negative speed. Speed decreases when velocity and acceleration have opposite signs; a negative acceleration alone does not tell you that speed decreases.
Read displacement-time and velocity-time graphs 4.2
The gradient of a displacement-time graph is velocity. A tangent gives instantaneous velocity; a chord between two times gives average velocity. A horizontal section means the particle is at rest.
The gradient of a velocity-time graph is acceleration. The signed area between that graph and the time axis is displacement. Areas below the axis count negatively; add absolute areas for distance travelled.
Worked example: velocity +4 m/s for 3 s then -2 m/s for 2 s gives displacement 4(3)-2(2)=8 m and distance 4(3)+2(2)=16 m. A vertical jump between these constant velocities is an idealisation, not finite acceleration over an interval.
Use constant-acceleration formulae only when valid 4.2
For constant acceleration a: v=u+at; s=ut+(1/2)at^2; s=vt-(1/2)at^2; v^2=u^2+2as; and s=(u+v)t/2. Here s is signed displacement and u and v are signed velocities. Choose equations containing the known quantities and the unknown you need.
Worked example: u=3 m/s, a=2 m/s^2 and t=4 s give v=11 m/s and s=28 m. The area under the velocity-time line is the same 28 m. The formulae do not apply to a varying acceleration over the whole interval.
A particle with u=12 m/s and a=-3 m/s^2 first stops at t=4 s. From 0=144-6s, its stopping displacement is 24 m. It has not reversed before this stop, so the distance is also 24 m. Do not keep using the stopping scenario after the stop unless the question says how it moves next.
Differentiate to find velocity and acceleration 4.2
For displacement s(t), v=ds/dt and a=dv/dt=d^2s/dt^2. For speed, take the magnitude of the velocity you find. Use techniques from Paper 1 and keep time units consistent.
Worked example: s=t^3-6t^2+9t gives v=3t^2-12t+9 and a=6t-12. At t=4, s=4 m, v=9 m/s and a=12 m/s^2. The average velocity on 0<=t<=4 is (4-0)/4=1 m/s, not the instantaneous value 9 m/s.
Solve v(t)=0 to locate possible changes in direction and check the sign on either side. In this example v=3(t-1)(t-3), so it is positive before t=1, negative between 1 and 3, and positive after 3. A zero velocity is not automatically a reversal; the sign check matters.
Integrate, use initial conditions and split at reversals 4.2
Integrating acceleration gives velocity plus a constant; integrating velocity gives displacement plus a constant. Use the given initial velocity and position to find each constant separately.
Worked example: a=6t-12 and v(0)=9 give v=3t^2-12t+9. Integrating with s(0)=0 gives s=t^3-6t^2+9t. A definite integral of v gives signed displacement without needing the displacement constant.
For that example, s(0)=0, s(1)=4, s(3)=0 and s(4)=4. Distance travelled on 0<=t<=4 is |4-0|+|0-4|+|4-0|=12 m; displacement is only 4 m. Split at times where velocity changes sign, or integrate the speed |v|.
Compare two particles with a common clock 4.2
Write a position equation for each particle using the same origin, positive direction and time reference. When they meet their positions are equal, not necessarily their displacements measured from different starts.
Worked example: A starts at the origin with u=2 m/s and a=1 m/s^2; B starts there at the same instant with constant velocity 6 m/s. Their positions are 2t+(1/2)t^2 and 6t. Equating gives t(t-8)=0. The later meeting is at t=8 s, position 48 m.
For a delayed start, use elapsed time t-d for the particle that starts d seconds later and enforce t>=d. Reject times outside the model's interval and distinguish the shared starting instant from a later meeting.
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