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?

A particle starts from origin att = 0 with a velocity 5.0 hati m/s and

1.	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?
Answer» Solution :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)`