A particle is travelling in a straight line. Its displacement, s metres, from the origin at time t seconds is given by s = 1.5e^(2t) + 2e^(-2t) - t.
- (a)Find expressions for the velocity, v ms⁻¹, and acceleration, a ms⁻², of the particle.[3]
- (b)Find the time, T seconds, when the particle is at rest.[4]
- (c)Find the acceleration of the particle at time T seconds.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [B2] B1 for 3e^(2t) or -4e^(-2t) for Velocity: 3e^(2t) - 4e^(-2t) - 1 isw
- [B1] FT me^(2t) + ne^(-2t) + k where m, n and k are constants for Acceleration: 6e^(2t) + 8e^(-2t) isw
- [B1] 3e^(4t) - e^(2t) - 4 = 0 or 3(e^(2t))² - e^(2t) - 4 = 0
- [M1] FT their 3-term quadratic in e^(2t) oe for (3e^(2t)-4)(e^(2t)+1) = 0
- [A1] e^(2t) = 4/3 nfww
- [A1] 1/2 ln(4/3) oe, isw or 0.144 or 0.1438[41...] rot to 4 or more dp and no other solutions
- [M1] FT pe^(2t) + qe^(-2t) where p and q are non-zero constants and their positive 1/2 ln(4/3) from part (b) for 6e^(2t) + 8e^(-2t) or 6(4/3) + 8(3/4)
- [A1] 14 nfww
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