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