A particle moves in the `x-y` plane with constant acceleration of `4 "m.s"^(-2)` in the direction making an angle of `60^(@)` with the x-axis. Initially, the particle is at the origin and its velocity is `5 "m.s"^(-1)` along the x-axis. Find the velocity and the position of the particle at `t = 5s.`

Explanation

According to the question if the initial velocity of the particl e along OA is u then (given `u_(x)= 5` m/s)
u = `u_(x)"cos"60^(@) = 5 xx (1)/(2) = (5)/(2) `m/s
The velocity of the particle at t = 5s,
v = u +at `= (5)/(2) +4xx5 `= 22.5 m/s

At that time the distance travelled by the particle along OA,
`s = ut +(1)/(2) at^(2) = (5)/(2)xx5+(1)/(2)xx4xx5^(2)`
`=(25)/(2)+50 = (125)/(2)` m
If the coordinate of the particle is (x,y) at that time then
x = `(125)/(2) "cos"60^(@)= (125)/(2)xx(1)/(2)= (125)/(4)`
and y = `(125)/(2) "sin"60^(@)= (125)/(2)xx(sqrt(3))/(2)= ,(125sqrt(3))/(4)`
`:.` Position of the particle at t = 5 s is `((125)/(4),(125sqrt(3))/(4))`.