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...

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 \(O\).