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The coordinate of a particle moving in ...

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

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 focus

D

The distance traveled by the particle in time interval `t=0" to "t=pi//2p` is a

Text Solution

Verified by Experts

The correct Answer is:
A, B
`x= a cos (pt), y=b sin (pt)`
Equation of path in x-y plane is `[x/a]^(2)+[y/b]^(2)=1`
i.e, the path of the particle is an ellipse. Position vector of a point P is `vecr=a cos p t hati+b sin pt hatj rArr vecv=p (-a sin pt hati+b cos pt hatj) and veca=p^(2) (-a cos pthati-b sin pt hatj)=-p^(2) vecr`
Accerleration directed towards the centre , Also `vecv. veca=0 at t=pi//2p`.
At `t=0, vecr=a hati, At t=pi//2p, vecr=b hatj`
So in time t =0 to `t=pi//2p` particle goes form A to B travelling a distance more than a.
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