The coordinates of a particle moving in a pla | vedclass

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Question

(d) $x = a\cos (pt)$ and $y = b\sin (pt)$ (given)

$	herefore $ $\cos pt = \frac{x}{a}$ and $\sin pt = \frac{y}{b}$

By squaring and adding

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Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(d) $x = a\cos (pt)$ and $y = b\sin (pt)$ (given)

$	herefore $ $\cos pt = \frac{x}{a}$ 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

${v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}(a\cos (pt)) = - ap\sin (pt)$

${v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}(b\sin (pt)) = bp\cos (pt)$

$	herefore \;\;\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}}} ight)\;\hat i + bp\cos p\left( {\frac{\pi }{{2p}}} ight)\hat j$$ = - ap\;\hat i$

Acceleration at $t = \frac{\pi }{{2p}}$

$\vec a = a{p^2}\cos p\left( {\frac{\pi }{{2p}}} ight)\;\hat i - b{p^2}\sin p\left( {\frac{\pi }{{2p}}} ight)\hat j$$ = - b{p^2}\hat j$

As $\vec v\;.\;\vec a = 0$

Hence velocity and acceleration are perpendicular to each other at $t = \frac{\pi }{{2p}}$.

The coordinates of a particle moving in a plane are given by $x = a\cos (pt)$ and $y(t) = b\sin (pt)$ where $a,\,\,b\,( < a)$ and $p$ are positive constants of appropriate dimensions. Then

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