A particle A, moving along a straight horizontal track with constant speed 8ms⁻¹, passes a fixed point O. Four seconds later, another particle B passes O, moving along a parallel track in the same direction as A. Particle B has speed 20ms⁻¹ when it passes O and has a constant deceleration of 2ms⁻². B comes to rest when it returns to O.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [M1] Use suvat to find expressions for SA or SB; For SA must be using u = 8 and time of t±4; For SB, using s = ut + ½at² with u = 20 and a = ±2
2. [A1] (SA=)8(4+t)[=32+8t]; Any unsimplified expression; ISW
3. [A1] (SB=)20t-½×2×t²; Any unsimplified expression; ISW; If 0 marks scored then allow SC: B1 for (SA =)8t and B1 for (SB =)20(t-4)-½×2×(t-4)² maximum 2/3
4. [*M1] Equating their expressions for SA and SB to form an equation in t; where SA is of the form ±8t±32 and SB is of the form ±20t±½×2×t²
5. [DM1] For reference t² −12t+32=0; Allow if no working seen and have correct real solution(s) to their 3-term quadratic. If working shown and if using the formula, it must be using the correct formula. If factorising must have 3 of the 4 terms correct of (t-4)(t-8)
6. [A1] t = 4 and 8; If 0 marks scored then allow SC: M1 for ±8t = ±20(t±4)±½×2×(t±4)² and A1 for t = 8 and 12 maximum 2/3.
7. [B1 FT] Straight line; Positive gradient, intersecting positive s axis. Full domain not required. FT if they get SA = 8t using the SC in (a)
8. [B1 FT] Inverted quadratic, passing through origin.; Full domain not required but must clearly go beyond the maximum. FT if they get SB = 20(t-4)-½×2×(t-4)² using the SC in (a), with curve though positive t axis before turning point.
9. [B1] All correct, line through (0, 32), quadratic through (20, 0), intersections indicated at t = 4 and t =8.; Intersections must occur before the turning point.