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})
= (51frac{1}{3} m)