A particle moves in a straight line. The particle is initially at rest at a point O on the line. At time t s after leaving O, the acceleration a m s¯² of the particle is given by a = 25 – t² for 0 ≤ t ≤ 9.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

1. [B1] Acceleration = 0 when t = 5 from 25 – t² = 0
2. [M1] Use of integration [v = 25t – ⅓t³]
3. [M1] Substitution for t [Max speed = 25 × 5 – ⅓ × 5³]
4. [A1] Max speed = 83⅓ m s¯¹
5. [M1] Use of integration [s = ½2t² – ¹/₁₂t⁴]
6. [A1] Distance = 260 m (260.4166...)
7. [B1ft] ft v from (i) At t = 9, v = 25 × 9 – ⅓ × 9³ = −18
8. [M1] Use of integration [ ∫ -3t¯½ dt = -6t½ ]
9. [M1] Substituting limits [Change in velocity from t = 9 to t = 25 = [-6t½]²⁵₉ = -6 × 5 + 6 × 3 = -12]
10. [A1] Velocity at t = 25 is –18 – 12 = –30 m s¯¹
11. [B1ft] ft v from (i) At t = 9, v = 25 × 9 – ⅓ × 9³ = -18
12. [M1] Use of integration [ ∫ −3t¯½ dt = −6t½ (+ C)]
13. [M1] Finds C and substitutes t = 25 [t = 9, v = −18 → C = 0, t = 25, v = −6 × 25½ ]
14. [A1] Velocity at t = 25 is –30 m s¯¹