A particle starts from origin att = 0 with a velocity `5.0 hati` 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-coordinate of the particle at the instant its x-coordinate 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)=5.0hati t+(1//2)(3.0 hati+2.0 hatj)t^(2)`
`=(5.0t+1.5t^(2))hati+1.0 t^(2) hatj`
Therefore `x(t)=5.0t+1.5t^(2), y(t)=+1.0 t^(2)`
Given `x(t)=84 m , t= `
`5.0t+1.5t^(2)=84 rArr t=6s`
At `t=s, y=1.0 (6)^(2)=36.0 m`
Now the velocity `v=(dr)/(dt)=(5.0+3.0t)hati+2.0 t hatj`
At `t=6 s, v=23. hatj+12.0 hatj`
Speed `=|v|= sqrt(23^(2)+12^(2))=26 m s^(-1)`