A particle P travels in a straight line, starting at rest from a point O. The acceleration of P at time t s after leaving O is denoted by a ms⁻², where a = 0.3t² for 0 ≤ t ≤ 4, a = -kt⁻³ᐟ² for 4 < t < T, where k and T are constants.

Mark scheme

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

1. [M1] Integrate acceleration for 0 ≤ t ≤ 4 to find velocity. v = ∫ 0.3t¹ᐟ² dt = 0.2t³ᐟ² (+c). The power of t must increase by 1 with a change of coefficient. (Note: The QP stated a = 0.3t², but the MS implies a = 0.3t¹ᐟ² for this calculation). Since P starts at rest from O, v(0)=0, so c=0.
2. [A1] Substitute t=4 into the velocity expression: v(4) = 0.2 × 4³ᐟ² = 0.2 × 8 = 1.6 ms⁻¹. (ISW any extra work using the second equation for a).
3. [*M1] Integrate acceleration a = -kt⁻³ᐟ² to find velocity for 4 < t ≤ T. v = ∫ -kt⁻³ᐟ² dt = 2kt⁻¹ᐟ² (+d). No need for constant 'd' at this stage. The power of t must increase by 1 with a change of coefficient. Use of v=at scores M0.
4. [A1 FT] Formulate two equations using the continuity condition at t=4 and the given velocity at t=16. For t=4: (Their 1.6) = 2k(4)⁻¹ᐟ² + d => 1.6 = k + d. For t=16: 0.3 = 2k(16)⁻¹ᐟ² + d => 0.3 = 0.5k + d. Allow unsimplified equations.
5. [DM1] Attempt to solve the two simultaneous equations for k or d. This involves correctly combining the equations to eliminate one variable.
6. [A1] Show that k = 2.6 (AG). And derive the expression for velocity: v = 5.2t⁻¹ᐟ² - 1. (Allow unsimplified expression for v and/or in terms of k, e.g., v = 2kt⁻¹ᐟ² - 1, and demonstrate that d = -1).
7. [M1] Set the velocity expression for 4 < t < T to zero to find T. Use their expression for v from part (b). e.g., 5.2T⁻¹ᐟ² - 1 = 0.
8. [A1] Find the exact value of T = 676/25 or 27.04. Must be exact. Allow if expression for v was correct in (b).
9. [M1] Integrate the velocity expressions from 0 to 4 and from 4 to T (their 27.04) to find displacement. For 0 ≤ t ≤ 4, integrate v₁ = 0.2t³ᐟ² (derived from a=0.3t¹ᐟ²). For 4 < t ≤ T, integrate v₂ = 5.2t⁻¹ᐟ² - 1. The power of t must increase by 1 with a change of coefficient.
10. [A1ft] Correctly integrate both parts to find displacement: ∫ (0.2t³ᐟ²) dt = 0.08t⁵ᐟ² and ∫ (5.2t⁻¹ᐟ² - 1) dt = 10.4t¹ᐟ² - t. Allow unsimplified, follow through on their velocity expression for v₂.
11. [M1] Apply correct limits for both integrals. For the first part, (0.08(4)⁵ᐟ² - 0.08(0)⁵ᐟ²). For the second part, (10.4(T)¹ᐟ² - T) - (10.4(4)¹ᐟ² - 4). Sum the two displacements. Use their value of T from (c).
12. [A1] Total distance = 64/5 or 12.8 (AWRT 12.8).