A particle starts from origin at t= 0 with a velocity `5.0hat(i)`m/s and moves in x-y plane under action of a force which produces a constant acceleration of `(3.0i+2.0j)"m/s"^(2)`. (a) What is the y-coordinates of the particle at the instant its x-coordinates is 84m? (b) what is the speed of the particle at this time?

Explanation

The position of the particle is given by
`r(t) =v_(0)t+(1)/(2)at^(2)=5.0hat(i)t+"(1/2)"(3.0hat(i)+2.0hat(j))t^(2)`
`=(5.0t+1.5t^(2))hat(i)+1.0t^(2)hat(j)`
Therefore, `x(t)=5.0t+1.5^(2), y(t)=+1.0t^(2)`
Given `X(t)=84m, t=?`
`5.0t+1.5t^(2)=84rArrt=6s`
At `t =6s, y=1.0(6)^(2)=36.0m`
Now, the velocity `v=(dr)/(dt)=(5.0+3.0t)hat(i)+2.0that(j)`
At t = 6s, `v=23.0hat(i)+12.0hat(j)`
speed `=|v|=sqrt(23^(2)+12^(2))=26ms^(-1)`