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) `vec(v) = u_(0) hat(i) + k omega cos omega t hat(j) = v_(x) hat(i) + v_(y) hat(j)` ,brgt `v_(x) = u_(0)`
`(d x)/(d t) = u_(0) rArr int_(0)^(x) dx = u_(0)int_(0)^(t) dt`
`x = u_(0) t` (i)
`rArr t = (x)/(u_(0))`
`v_(y) = k omega cos omega t`
`(d y)/(d t) = k omega cos omega t rArr int_(0)^(y) dy = k omega int_(0)^(t) cos omega t dt`
`y = k omega|(sin omega t)/(omega)|_(0)^(t) = k sin omega t` (ii)
Eliminating `t` from (i) and (ii), we get
`y = k sin omega t = k sin ((omega t)/(u_(0)))`
`= k sin((omega x)/(u_(0))) ["sine curve"]`
If `(omega x)/(u_(0)) = (pi)/(2) rArr x = (pi u_(0))/(2 omega), y = k`
`(omega x)/(u_(0)) = pi rArr (pi u_(0))/(omega), y = 0`

At `t = (3pi)/(2 omega), x = u_(0) t = (3 pi u_(0))/(2 omega)`
`y = k sin ((omega. 3pi)/(2omega)) = - k`
`r = sqrt(x^(2) + y^(2)) = sqrt(((3pi u_(0))/(2 omega))^(2) + k^(2))`
where `r` is the distance from the origin.