A-LevelMathematicsMechanics9709/41 · May/June 2017 · Q610 Marks
A particle P moves in a straight line passing through a point O. At time t s, the velocity of P, v m s⁻¹, is given by v = qt + rt², where q and r are constants. The particle has velocity 4 m s⁻¹ when t = 1 and when t = 2.
- (i)Show that, when t = 0.5, the acceleration of P is 4 m s⁻².[4]
- (ii)Find the values of t when P is at instantaneous rest.[2]
- (iii)The particle is at O when t = 3. Find the distance of P from O when t = 0.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [q + r = 4 and 2q + 4r = 4]
- q = 6 and r = –2 so v = 6t – 2t²
- a = 6 – 4t
- At t = 0.5, a = 4
- v = 6t – 2t² = 0
- t = 0 and t = 3
- EITHER: (M1 Attempt to integrate v to find s)
- s = ∫(6t - 2t²) dt
- s = 3t² – (2/3)t³ + C
- 0 = 3 × 3² – (2/3) × 3³ + C (M1 Use s = 0 when t = 3 to find C)
- C = –9 so distance = 9 m (A1) Valid argument
- OR: (M1 Attempt integration with limits)
- S = ∫₀³ (6t-2t²)dt
- Correct integration and correct limits but no evaluation
- [27 – 18 = 9] (M1 Evaluation of integral between limits)
- Distance from O at t = 0 is 9 m (A1) With explanation
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →