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 `(3.0hat(i) + 2.0hat(j)) 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 + 1/2at^(2)`
`=5.0hatit + (1//2) (3.0hati + 2.0hatj)t^(2)`
Therefore, `x(t) =5.0t + 1.5t^(2)`
`y(t) = +1.0t^(2)`
Given `x(t) =84 m, t=?`
`5.0t + 1.5t^(2) =84 rArr t=6s`
at t=6s, y=1.0 `(6)^(2)=36.0`m
Now, the velocity `v=(dr)/(dt) =(5.0 +3.0t)hati + 2.0hatj`
At t=6s, `v=23.0hati + 12.0 hatj`
speed =|v| `=sqrt(23^(2) + 12^(2)) =26 ms^(-1)`