A particle P starts from a point O and moves in a straight line. The velocity v ms⁻¹ of P, at time t s after leaving O, is given by v = ¹/₃(2t-3)(t-4).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [M1] Expand to a 3-term quadratic equivalent and attempt to differentiate (oe e.g. product rule), decrease power by 1 and a change in coefficient in at least one term; allow un-simplified a = v/t is M0.
2. [A1] ¹/₃(4×2-11)=-1m s⁻² Must be negative.
3. [*M1] Attempt to integrate Expand to a 3-term quadratic equivalent and attempt to integrate, increase power by 1 and a change in coefficient on the same term. s = vt is M0.
4. [A1] s = ¹/₃(2t³/₃ - ¹¹t²/₂ + 12t) (+c) Allow un-simplified.
5. [*B1] [P changes direction when] t = 1.5 SOI.
6. [DM1] Attempt to evaluate their ¹/₃(2t³/₃ - ¹¹t²/₂ + 12t) for t = 0 to t = 1.5 and t = 1.5 to t = 3 Using the correct limits correctly.
7. [A1] Total distance is [2.625+|–1.125|=] 3.75 m If no integration seen, then B1 for 3.75 (so 2 marks max.).
8. [M1] Attempt to evaluate their s for t = 0 to t = 4 OE e.g. compare ∫_0^{1.5} vdt with ∫_{1.5}^4 vdt or setting their three-term cubic expression in t (with no constant term) equal to zero and attempting to solve or attempt discriminant.
9. [A1] ∫_0^4 vdt = ⁸/₉ > 0 so, no P does not return to O OE e.g. obtaining 4t² -33t +72 = 0 and showing that this has no real roots (and hence P does not return to O). 4t²-33t+72=0 gives b²-4ac = (-33)²-4×4×72=-63. 2/9 t²-11/6 t+4=0 gives b²-4ac = (-11/6)²-4×2/9×4=-7/36.