The coordinate of a particle moving in a plane are given by ` x(t) = a cos (pt) and y(t) = b sin (pt)` where `a,b (lt a)` and `P` are positive constants of appropriate dimensions . Then

Options

A. The path of the particle is an ellipse
B. The velocity and acceleration of the particle are normal to each other at `t=pi/(2p)`
C. The acceleration of the particle is always directed towards a fixed position
D. The distance travelled by the particle in time internal `t=0` to `t=pi/(2p)` is `a`

Answer

A, B, C

Explanation

`x^(2)/a^(2)+y^(2)/b^(2)=1`,So the path is an ellipse
`V_(x)=-ap sin pt,V_(y)=bp cos pt`
`a_(x)=-ap^(2) cos pt a_(y)=-bp^(2) sin pt`
So `vecV-veca=0` when `vecV_|_veca`
So `a^(2)p^(3) sin pt. cos pt-b^(2) p^(3) sin pt. cos pt`
`rArr a^(2)p^(3) sin pt. cos pt-b^(2) p^(3) sin pt. cos pt`
as`a ne b ` So `sin` pt. `cos pt=0`
`rArr sin p2t=0 rArr ,2pi rArr t=pi/(2P)--`
The motion is similar to motion of earth around sun So force always towards focus and hence acceleration. At `t=0` particle is at `(a,o)`
At `t=pi/(2p)` particle is at `(o,b)`
So distance travelled along `X` axis is a not the actual distance, which is the length of the part of the elipse between `(a,o)` to `(o,b)` you can try out for distance by following method `ds=sqrt(dx^(2)+dy^(2))`
`rArr ds=sqrt(1+((dy)/(dx))^(2)) rArr int_(0)^(s)=int_(a)^(0)sqrt(1+((dy)/(dx))^(2)).(dx)`