A particle P moves in a straight line. The velocity vms⁻¹ at time t seconds is given by v = 0.5t for 0≤t≤ 10, v = 0.25t² – 8t + 60 for 10 ≤ t ≤ 20.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [B1] (a =) dv/dt = 0.5; For acceleration during the first 10 seconds
2. [B1] Differentiate to get (a =) dv/dt = 2×0.25t-8 [=0.5t-8]; Allow unsimplified
3. [B1] a[=0.5×10-8]=-3; CWO. Do not award final B mark if more than 2 accelerations seen and not discarded, 2/3 maximum; Ignore any comments, correct or incorrect
4. [B1] Get distance in first 10 seconds as 25; From suvat or from ∫(0.5t)dt from 0 to 10
5. [B1] v = 0 when t=12 and t = 20; SOI
6. [*M1] Attempt to integrate v; For integration, the power of t must increase by 1 in at least 1 term with a change of coefficient in the same term. s = vt is M0; s = ∫(0.25t² - 8t + 60)dt = 0.25t³/3 - 8t²/2 + 60t (+c)
7. [A1] s = 0.25t³/3 - 4t² + 60t (+c) [= t³/12 - 4t² + 60t (+c)]; Allow unsimplified
8. [DM1] Attempt to evaluate their [t³/12 - 4t² + 60t] for t = 10 to t = 12 and t = 12 to t = 20; Using the correct limits correctly
9. [A1] s = [25 + 288 - (850/3 - 800/3) - 288 + (14/3 - 64/3)] = 51 m