A-LevelMathematicsMechanics9709/43 · May/June 2022 · Q710 Marks
A particle P moves in a straight line through a point O. The velocity vms⁻¹ of P, at time t s after passing O, is given by `v = 9/4 + b / (t+1)² - ct²`, where b and c are positive constants. At t = 5, the velocity of P is zero and its acceleration is -13/12 ms⁻².
- (a)Show that b = 9 and find the value of c.[5]
- (b)Given that the velocity of P is zero only at t = 5, find the distance travelled in the first 10 seconds of motion.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [B1] Use of v(5) = 0 to form equation in b and c
- [M1] For use of a = dv/dt and a(5) = -13/12
- [A1] a = -2b/(t+1)³ - 2ct; -13/12 = -2b/6³ - 2c(5)
- [M1] Attempts to solve simultaneous equations
- [A1] b = 9 and c = 0.1 (AG)
- [M1] For use of s = ∫v dt
- [A1 FT] s = (9/4)t - 9/(t+1) - 0.1t³/3 + K [FT their value of c from (a)]
- [M1] For evaluation from 0 to 5 or from 5 to 10
- [M1] For evaluation from 0 to 5 and from 5 to 10 to find distance travelled
- [A1] = 31.8 m or 350/11
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