A particle is moving in a plane with velocity `vec(v) = u_(0)hat(i) + k omega cos omega t hat(j)`. If the particle is at origin at `t = 0`, (a) determine the trajectory of the particle. (b) Find its distance from the origin at `t = 3pi//2 omega`.

Options

A. `sqrt([((3piu_(0))/(2 omega))^2+a^2])`
B. `sqrt([((3pi u_0)/(2 omega))+a^2])`
C. `sqrt([((3piu_0)/(2omega))^2+a]]`
D. `sqrt([((4piu_0)/(2omega))^2+a^2])`

Explanation

Let `u_(x)` and `u_(y)` be the components of the velocity of the particle along the x- and y- directions . Then `u_(x)=dx//dt` and `u_(y)=dy//dt= omega` cos `omegat` Intergration : `X = u_(0)` t and y =a `sin omegat`
Elimination , `y=a sin (omega x//u_(0))`
This is the equation of the trajectory
At `t=3 pi //2 omega` , We have , `x=u_(0) 3 pi//2 omega` and `y = a "sin"3 pi//2=-a`
`therefore` The distance of the particle from the origin is
`sqrt(x^2+y^2)=sqrt([((3 pi u_0)/(2 omega))^2+a^2]`