A particle moves according to the law, `x=acos(pit//2).` . What is the distance covered by it in time interval `t=0` to `t=3` second.

Options

A. 2a
B. 3a
C. 4a
D. a

Answer

2

Explanation

`X=ncoswt`
`omega=(pi)/(2)`
`(2pi)/(T)=(pi)/(2)`
`T=4sec`

The given time t=3 sec is really `=(3T)/(4)`
At t=0 particle is at x=a (at extreme postion ) & at t = 3s.
i.e. `(3T)/(4)` it is at means position so distance =3a.