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

Explanation

The position of the particle is given by
`vecr(t)=vecV_(0)t+1/2vecat^(2)`
`=5hatit+(1//2)(3hati+2hatj)t^(2)`
`=(t+1.5t^(2))hati+t^(2)hatj`
Therefore, `x(t)=5t+1.5t^(2),y(t)=t^(2)`
`5t+1.5t^(2)=84rArr t=6s`
At `t=6s,y=(6)^(2)=36 m`
`vecv=(dvecr)/dt=(5+3 t)hati+2t hatj`
At `t=6s,vecv=23hati+12hatj`
speed `=|vecv|=sqrt(23^(2)+12^(12))=26ms^(-1)`