A particle starts from rest at a point O. The acceleration of the particle at time t s after leaving O is a m s⁻², where a = 2(t+1)⁻½ - 1 for t ≥ 0. Find the distance that the particle travels from O until the time at which its acceleration is zero.

Mark scheme

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

1. [B1] Acceleration = 0 when 2(t+1)⁻½ - 1 = 0 ⇒ (t+1)½ = 2 ⇒ t=3; CWO. Those who use 2(t+1)⁻½ or claim that 2(t+1)⁻½-1 is (2t+2)⁻½-1 score 0 marks.
2. [*M1] For attempt at integration of (a =)2(t+1)⁻½-1; Of the form a(t+1)½ -t[+c], where a≠2.
3. [A1] [v=] -(t+1)½+ -t[+c] = 4(t+1)½-t[+c]; Allow un-simplified.
4. [DM1] v=0 at t=0 ⇒ c=-4 ⇒ v=4(t+1)½-t-4; For attempt to find c using v=0 at t=0. Must get a value for c. M0 for stating v=0at t=0⇒c=0 with no working.
5. [DM1] Attempt to integrate their v, which has come from integration; Increase power by 1 and change of coefficient on the same term from a 3 term expression in t to get an expression of the form p(t+1)³/² + qt² + rt, where p≠0, q≠0 and r≠0. s=vt_is M0. Dependent on previous 2 M marks.
6. [A1] [s=] (t+1)³/² - ½t² - 4t[+k] = (t+1)³/² - ½t² - 4t[+k]; Allow un-simplified. Condone using +c again.
7. [DM1] OR evaluate k from using s=0 when t=0 AND substitute t = their3; Use of limits 0 and their positive 3 correctly, their 3 must have come from considering given a=0. Dependent on all 3 previous M marks. If evaluating k, then k = 8/3 if correct.
8. [A1] Distance = 13/6 m or 2.17m; SC, If no integrating v seen then SCB1 for 2.17 so B1M1A1M1SCB1 5 marks max