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`.

Explanation

(a) Here, ` vec u =u_x hati + u_y hat j + ( omega acos omega t) hat j`
` S0, u_x =u-o and u_y = omega cos omega t`
` u_x = 9dx)/(dt0 =u_0 or dx = u_0 dt`
Integrating it `, we have ltBrgt ` x =u_0 t or t= x//u_0`
Aslso ,` u_y = (dy) /dx) = omega a cos omega t`
` or ` dy = omega a cos omega t dt`
Integarating it, we have
y= omega a ( sin omega t)/(omega) = a sin omega t= a sin omega x// u_0`
Thus equation of trajectory is
` y= a sin omega x // u_0`
(b) When ` t= 3 pi //2 omega , x= u_0 (3 pi) /( 2 omega) ` and
` y= a sin omega xx (3 pi) /(2 omega) = a sin ( 3 pi)/ 2 = a(-1) =- a`
Distanc from origin
` r sqrt ( x^2 = y^2 ) = sqrt (( u_0 3 pi // 2 omega)^2 + (-a)^2)`
` = sqrt (9 pi^2 u_0^2)/(4 omega) + a^2 )`.