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

Explanation

From Eq . (4.34a) for `r_(0)=0`, the position of the particle is given by `r(t)=v_(0)t+(1)/(2)at^(2)`
`=5.0hati+(1//2)(3.0hati+2.0hatj)t^(2)`
Therefore , `x(t)=5.0t+1.5t^(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+3.0)hati+2.0hatj`
At `t=6s,v=23.0hati+12.0hatj`
speed `=|v|=sqrt(23^(2)+12^(2))~=26 " m s"^(-1)`.