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.
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.
\(s = 2(5^{2}) - \frac{1}{3} (5^{3}) + 5(5) + 18\)
= \(50 - \frac{125}{3} + 25 + 18\)
= \(\frac{154}{3}\)
= \(51\frac{1}{3} m\)
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