A particle P moves in a straight line. At time ts after passing through a point O on the line, the displacement of P from O is sm, where s = 0.01t³ − 0.3t² − 2.07t.

Mark scheme

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

1. [1] Attempt to differentiate s=0.01t³ – 0.3t² – 2.07t. Decrease power by 1 and change of coefficient on the same term; a = v/t is M0.
2. [1] [v =] 0.03t² – 0.6t – 2.07 Allow un-simplified.
3. [1] [a =] 0.06t – 0.6 Allow un-simplified. Must follow from the correct 3 term quadratic in t for v so A0 if coming from v = 0.03t² – 0.6t. Or the correct completing the square form for the correct expression for v.
4. [1] Setting their a = 0 where a is a two-term linear expression (of the form k₁t+k₂ where k1,k2 ≠ 0) in t. Or setting their t – 10 (from completing the square on a three term quadratic in t) equal to 0.
5. [1] s = -40.7 CWO (so first four marks must have been awarded). Must be negative so do not allow s = -40.7 ⇒ s = 40.7 so no ISW. Saying that s = –40.7 or 40.7 is A0.
6. [1] v = -5.07 ⇒ speed is 5.07 m s⁻¹ CWO (so first four marks must have been awarded or first three marks if completing the square). Must be positive – allow A1 if negative sign is dropped without justification but saying speed is either -5.07 or 5.07 is A0.
7. [1] Set their 3 term quadratic expression in t for v equal to zero - dependent on first M1 mark in 4(a). This mark can be implied by a correct value of t = 23 appearing in this part.
8. [1] t = 23 CWO Ignore mention of t = – 3. Working might appear in 4(a), but this value must be stated in 4(b) to earn this mark.
9. [1] Their positive t substituted into their two-term linear expression for a - this mark can be implied by a correct answer.
10. [1] a = 0.78 m s⁻² ONLY OE e.g. 39/50 etc. if -0.78 found then must be explicitly rejected.