A particle P moves in a straight line and passes through the point A at time t = 0. The velocity vms⁻¹ of P at time t seconds is given by v = (2t+1)³⁄₂ – 2t², where 0 ≤ t ≤ 3.

Mark scheme

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

1. [1] (a) Attempt to differentiate (v=)(2t+1)³⁄₂ – 2t² Decrease power by 1 and a change in coefficient in at least one term; a=v/t is M0.
2. [1] (a) dv/dt = 3/2(2t+1)¹⁄₂ ×2–2×2t¹ [= 3(2t+1)¹⁄₂–4t] Allow un-simplified.
3. [1] (a) dv/dt=0 ⇒ 3(2t+1)¹⁄₂–4t = 0 ⇒ 9(2t+1)=16t² Setting their derivative (of the form p(2t+1)¹⁄₂–qt where p≠0,q≠0) equal to zero and uses a correct method to form an equation without square roots involved, which would lead to a quadratic in t.
4. [1] (a) 16t²-18t-9=0⇒t=... Solving correctly their three-term quadratic in t (must be using correct formula OR if factorising, two terms must be a correct multiple of their three-term quadratic when expanding brackets e.g. 16t² -18t-9=0 followed by (t-3/2)(t+3/8)=0 is OK for M1), to obtain (at least) one positive value of t. Must get a value of t such that 0<t<3. For reference t = 1.5 (1 - 3/8). Dependent on both previous M marks. If no method seen, must have at least one correct positive value for their three-term quadratic for this mark.
5. [1] (a) Maximum velocity is 3.5 m s⁻¹ ONLY CWO.
6. [1] (b) Attempt to integrate (v =)(2t+1)³⁄₂ – 2t² Increase power by 1 and a change in coefficient in at least one term; s = vt is M0.
7. [1] (b) (s =) (2t+1)⁵⁄₂/(5/2) × 1/2 – 2t³⁄3 (+c) Allow un-simplified.
8. [1] (b) Substitute t = 1.5 and t = 0 OR evaluate c(≠0) using t = 0 and substitute t = 1.5 Use of t = 0 and their 1.5 correctly (in their integrated expression) ONLY; F(their1.5)-F(0); their 1.5 must have come from an attempt at differentiation in (a) hence dependent on the first M1 in (a). Where their t is such that 0<t<3. For reference c = -0.2.
9. [1] (b) Distance = [ (32/5) – (9/4) ] – 0 = 3.95 or 79/20 m CWO. If no integration seen allow SCB1 for 3.95. For reference c = -0.2.