A particle X moves in a straight line. The displacement of X from O at time ts after leaving O is sm, where s = 0.3t² +0.6t for 0≤t≤4.

Mark scheme

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

1. [B1] v=0.6t+0.6 Allow un-simplified.
2. [B1] Velocity at t = 4 is 3 m s⁻¹
3. [*M1] For attempt at integration of 0.3t^2 to kt^3 (+c) Answer must be of the form kt^3 where k≠0 or 0.3 Use of v = at or v = k₁ (t −4)^2 + ... scores M0. Do not penalise missing c.
4. [DM1] Their 3 = 0.2(4)^2 + c leading to c=... ⇒v=0.2t^2+1.4 Use of their3 from part (a) and t = 4 to find c (for reference if correct then c = 1.4) OR using correct limits of 4 and T in their equation for v e.g. [0.2t^2]_4^T = 0.2(T^2-8).
5. [DM1] 14.2=0.2T^2 + their 1.4 For equating to 14.2. Allow in terms of t Dependent on both previous M marks OR equivalent e.g. 14.2 – their 3 = 0.2[T^2-8].
6. [A1] T=16 Allow 16 or t = 16.
7. [B1] Distance travelled in first 4 seconds = 7.2m
8. [*B1FT] s = 0.08t^3 + 1.4t For integrating their v from part (b)(i) correctly which must be of the form αt² + β with α, β ≠ 0 – allow un-simplified.
9. [DM1] (0.08(16)^3 + 1.4×16) - (0.08(4)^3 + 1.4×4) [=81.92+22.4-2.56-5.6=104.32-8.16=96.16] Correct use of limits 4 and their 16 (> 4) must be equivalent to F(their 16) – F(4).
10. [A1] Distance = 103.36 m Condone 103 or better CWO, 2584/25. Do not ISW if, for example, 103.36 + 7.2 = 110.56. If constant of integration c found, then must be correct, that is c = -0.96. If no integration seen than max B1 for 7.2 and B1 for 103 m or better.