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
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
The path of the particle is an ellipse
The velocity and acceleration of the particle are normal to each other at $t = \pi /(2p)$
The acceleration of the particle is always directed towards a focus
Both $(a)$ and $(b)$
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
$x = a\cos (pt)$ 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 $t = \frac{\pi }{{2p}}$.
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