Answer: (a) \(20\cos2t+12\sin2t\); (b) \(12\text{ m s}^{-2}\); (c) \(t\approx1.06\); (d) \(8\text{ m}\).

We start with the main method. Displacement is found by integrating velocity over the given time interval.

(a) The velocity is

\(v=10\sin2t-6\cos2t.\)

Differentiate with respect to \(t\):

\(a=20\cos2t+12\sin2t.\)

(b) When \(t=\frac\pi4\), \(2t=\frac\pi2\). Therefore

\(a=20\cos\frac\pi2+12\sin\frac\pi2.\)

So

\(a=0+12=12.\)

(c) The acceleration is zero when

\(20\cos2t+12\sin2t=0.\)

Thus

\(12\sin2t=-20\cos2t.\)

So

\(\tan2t=-\frac{20}{12}=-\frac53.\)

The first positive solution occurs in the second quadrant for \(2t\):

\(2t=\pi-\tan^{-1}\frac53.\)

Therefore

\(t=\frac12\left(\pi-\tan^{-1}\frac53\right).\)

Numerically,

\(t\approx1.06.\)

(d) Displacement is found by integrating velocity:

\(\int(10\sin2t-6\cos2t)\,\mathrm dt=-5\cos2t-3\sin2t+c.\)

So the displacement between \(t=\frac\pi4\) and \(t=\frac\pi2\) is

\(\left[-5\cos2t-3\sin2t\right]_{\pi/4}^{\pi/2}.\)

At \(t=\frac\pi2\), this is

\(-5\cos\pi-3\sin\pi=5.\)

At \(t=\frac\pi4\), this is

\(-5\cos\frac\pi2-3\sin\frac\pi2=-3.\)

Therefore the displacement is

\(5-(-3)=8.\)

This completes the solution and gives the required result.