The velocity of a particle at time t s after leaving a fixed point O is vms¯¹. The diagram shows a velocity-time graph which models the motion of the particle. The graph consists of 5 straight line segments. The particle accelerates to a speed of 0.9ms¯¹ in a period of 3s, then travels at constant speed for 6 s, and then comes instantaneously to rest 1 s later. The particle then moves back and returns to rest at O at time T s. [Figure 4.1]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [M1] Completely correct method for finding the total area underneath the velocity-time graph from t = 0 to 10 only. Can be done as two triangles and a rectangle e.g., ¹/₂×3×0.9+(9-3)×0.9+¹/₂×1×0.9 (allow a slip in one value); need not see all three components added together.
2. [A1]
3. [M1] Setting ¹/₂(12-10)v_min equal to ± their (a).
4. [A1] Must be negative – allow those who solve ¹/₂(12-10)v_min = 7.2 and obtain v_min = 7.2 and then change to -7.2 without justification. SC B1 for assuming the triangle is isosceles.
5. [*M1] Correct method for finding T or t. Condone sign errors but must equate to their answer to (a).
6. [A1] OE (e.g. from t + 10 = 4.8 + 10 = 14.8).
7. [DM1] M1 for 2(their (a)) / their T or 2(their (a)) / (10+ their T)
8. [A1] OE 0.973 [For reference: 0.97297...]. SC *B1 (for T = 14.8). DM1A1 for assuming the triangle is isosceles.