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.

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A particle moves from point O along a straight line such that its acceleration at any time, t seconds is

a = (4 - 2 t) m s^{- 2}

. At t = 0, its distance from O is 18 metres while its velocity is

5 m s^{- 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 - 2 t) m s^{- 2}

∴ v = \int a d t

\int (4 - 2 t) d t = (4 t - t^{2} + c)

When t = 0, v = 5 m/s

5 = 0 - 0 + c ⟹ c = 5

∴ v = (4 t - t^{2} + 5) m s^{- 1}

Distance,

s = \int (4 t - t^{2} + 5) d t

2 t^{2} - \frac{t^{3}}{3} + 5 t + c

When t = 0, s = 18 m.

0 - 0 + 0 + c = 18 ⟹ c = 18

∴ s = 2 t^{2} - \frac{t^{3}}{3} + 5 t + 18

(a) Velocity is greatest when a = 0.

4 - 2 t = 0 ⟹ t = 2 s

(b)(i) Particle is momentarily at rest when v = 0 m/s.

4 t - t^{2} + 5 = 0 ⟹ t^{2} - 4 t - 5 = 0

t (t - 5) + 1 (t - 5) = 0 ⟹ 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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