d
(d) \(x = a\cos (pt)\) and \(y = b\sin (pt)\) (given)

\(\therefore \) \(\cos pt = \frac{x}{a}\) and \(\sin pt = \frac{y}{b}\)

By squaring and adding

\({\cos ^2}(pt) + {\sin ^2}(pt) = \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\)

Hence path of the particle is ellipse.

Now differentiating \(x\) and \(y\) w.r.t. time

\({v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}(a\cos (pt)) = - ap\sin (pt)\)

\({v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}(b\sin (pt)) = bp\cos (pt)\)

\(\therefore \;\;\vec v = {v_x}\hat i + {v_y}\hat j = - ap\sin (pt)\hat i + bp\cos (pt)\hat j\)

Acceleration \(\vec a = \frac{{d\vec v}}{{dt}} = \frac{d}{{dt}}[ - ap\sin (pt)\hat i + bp\cos (pt)\hat j]\)

\(\vec a = - a{p^2}\cos (pt)\;\hat i - b{p^2}\sin (pt)\hat j\)

Velocity at \(t = \frac{\pi }{{2p}}\)

\(\vec v = - ap\sin p\left( {\frac{\pi }{{2p}}} \right)\;\hat i + bp\cos p\left( {\frac{\pi }{{2p}}} \right)\hat j\)\( = - ap\;\hat i\)

Acceleration at \(t = \frac{\pi }{{2p}}\)

\(\vec a = a{p^2}\cos p\left( {\frac{\pi }{{2p}}} \right)\;\hat i - b{p^2}\sin p\left( {\frac{\pi }{{2p}}} \right)\hat j\)\( = - b{p^2}\hat j\)

As \(\vec v\;.\;\vec a = 0\)

Hence velocity and acceleration are perpendicular to each other at \(t = \frac{\pi }{{2p}}\).