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

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The path of the particle is an ellipse.

The velocity and acceleration of the particle are normal to each other at $t = \frac{π}{2 p}$

The acceleration of the particle is always directed towards a fixed position

The distance travelled by the particle in time internal $t = 0 t o t = \frac{π}{2 p}$ is ‘a’

answer is A, B, C.

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Detailed Solution

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ ; So the path is an ellipse

$V_{x} = - a p \sin p t, V_{y} = b p \cos p t a_{x} = - a p^{2} \cos p t, a_{y} = - b p^{2} \sin p t S o, \vec{V} . \vec{a} = 0 w h e n \vec{V} ⊥ \vec{a} So, a^{2} p^{3} \sin p t . \cos p t - b^{2} p^{3} \sin p t . \cos p t \Rightarrow a^{2} p^{3} \sin p t . \cos p t = b^{2} p^{3} \sin p t . \cos p t as a \neq b So \sin p t . \cos p t = 0 \Rightarrow \sin p 2 t = 0 \Rightarrow p 2 t = π, 2 π \Rightarrow t = \frac{π}{2 p} - - - - - N o w \vec{a} = - a p^{2} \cos p t \hat{i} - b p^{2} \sin p t \hat{j} = - p^{2} \vec{r}$

From the above equation it is clear that $\vec{a} a n d \vec{r} a r e o p p o s i t e l y d i r e c t e d a n d s o a c c e l e r a t i o n i s a l w a y s d i r e c t e d t o w a r d s t h e o r i g i n .$

$d s = \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x \Rightarrow \int_{0}^{s} d s = \int_{a}^{0} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ = Cicumference of the elliptical path which is greater than a.