A particle moves from point O along a straight line such that its acceleration at any time, t seconds is \(a = (4 - 2t) ms^{-2}\). At t = 0, its distance from O is 18 metres while its velocity is \(5 ms^{-1}\). (a) At what time will the velocity be greatest? (b) Calculate the : (i) time ; (ii) distance of the particle from O when the particle is momentarily at rest.
Explanation
Acceleration, \(a = (4 - 2t) ms^{-2}\) \(\therefore v = \int a \mathrm {d} t \) = \(\int (4 - 2t) \mathrm {d} t = (4t - t^{2} + c)\) When t = 0, v = 5 m/s \(5 = 0 - 0 + c \implies c = 5\) \(\therefore v = (4t - t^{2} + 5) ms^{-1}\) Distance, \(s = \int (4t - t^{2} + 5) \mathrm {d} t\) = \(2t^{2} - \frac{t^{3}}{3} + 5t + c\) When t = 0, s = 18 m. \(0 - 0 + 0 + c = 18 \implies c = 18\) \(\therefore s = 2t^{2} - \frac{t^{3}}{3} + 5t + 18\) (a) Velocity is greatest when a = 0. \(4 - 2t = 0 \implies t = 2s\) (b)(i) Particle is momentarily at rest when v = 0 m/s. \(4t - t^{2} + 5 = 0 \implies t^{2} - 4t - 5 = 0\) \(t(t - 5) + 1(t - 5) = 0 \implies t = -1 ; t = 5\) Since time cannot be negative, t = 5s. (ii) Distance when t = 5s. \(s = 2(5^{2}) - \frac{1}{3} (5^{3}) + 5(5) + 18\) = \(50 - \frac{125}{3} + 25 + 18\) = \(\frac{154}{3}\) = \(51\frac{1}{3} m\)