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 fr...
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A particle moves from point O along a straight line such that its acceleration at any time, t seconds is \(a = (4 - 2t) ms^{-2}\). 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.
Answer Details
Integrate the acceleration to get velocity, then position, applying the initial conditions.
\[v = \int (4 - 2t)\,dt = 4t - t^2 + C.\]
At \(t = 0,\ v = 5\), so \(C = 5\) and \(v = 4t - t^2 + 5\).
(a) Velocity is greatest when \(\dfrac{dv}{dt} = a = 0\):
\[4 - 2t = 0 \Rightarrow t = 2\ \text{s}.\]
(b) The particle is momentarily at rest when \(v = 0\):
\[4t - t^2 + 5 = 0 \Rightarrow t^2 - 4t - 5 = 0 \Rightarrow (t-5)(t+1) = 0 \Rightarrow t = 5\ \text{s}\ (t>0).\]
(i) Time \(= 5\ \text{s}\).
Now integrate for displacement: \(s = \int(4t - t^2 + 5)\,dt = 2t^2 - \dfrac{t^3}{3} + 5t + D\). At \(t = 0,\ s = 18\), so \(D = 18\).
(ii) At \(t = 5\):
\[s = 2(25) - \frac{125}{3} + 25 + 18 = 93 - \frac{125}{3} = \frac{154}{3} \approx 51.3\ \text{m}.\]
The particle is \(\dfrac{154}{3}\ \text{m} \approx 51.3\ \text{m}\) from \(O\).
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