A particle P moves in a straight line, starting from rest at a point O on the line. At time t s after leaving O the acceleration of P is k(16 – t²) m s⁻², where k is a positive constant, and the displacement from O is s m. The velocity of P is 8 m s⁻¹ when t = 4.
- (a)Show that s = ¹⁄₆₄t²(96 – t²).[5]
- (b)Find the speed of P at the instant that it returns to O.[3]
- (c)Find the maximum displacement of the particle from O.[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [M1] Uses v = ∫a dt. (a = 16k - kt², v = 16kt - ⅓kt³)
- [M1] Substitutes t = 4, v = 8. (8 = 16k × 4 - ⅓k × 4³ leading to k = ...)
- [A1] OE (v = 16kt - ⅓kt³ and k = 3/16)
- [M1] Uses s = ∫vdt and attempts to find s in terms of t only. May be using v = 3t - 1/16t³.
- [A1] AG, no errors seen. (s = 1/64t²(96 - t²))
- [M1] Attempt to find t when s = 0. (s = 0, t²(96 - t²) = 0, t = 4√6)
- [M1] Attempt to find v at this t value
- [A1] Do not condone v = -29.4. (Speed is 29.4 m s⁻¹)
- [M1] Determine the time, t (or t²) at which v = 0 (v = 0, t² = 48, t = 4√3)
- [M1] Use substitution of the t or t² value to find s.
- [A1] s = 36 m
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