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

In terms of components :
`vec(v) = hat(i) v_(x) + hat(j) v_(y)`
So, `v_(x) = u_(0)` and `v_(y) = a omega cos omega t`
i.e., `(dx)/(dt) = u_(0)` and `(dy)/(dt) = a omega cos omega t`
Which on integration gives
`x = int u_(0) dt` and `y = int a omega cos omega t dt`
i.e., `x = u_(0) t + A` and `y = a sin omega t + B`
Now as at `t = 0, x = 0` and `y = 0`
So, `A = 0` and `B = 0`
`:. x = u_(0)t`....(i) and `y = a sin omega t`
(a) Substituting the value of `t` from Eqn. (i) in (ii), `y = a sin (omega x//u_(0))`
Which is the desired trajectory and is a since curve as shown in Fig. 4.9.

(b) For `t = (3 pi //2 omega)` from Eqns. (i) and (ii), we have i.e., `x = u_(0) (3pi//2 omega)` and `y = - a`
So, distance from the origin at `t = (3pi//2omega)` will be
`r = sqrt(x^(2) + y^(2)) = sqrt([a^(2) + (3pi u_(0)//2 omega)^(2)])`