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 (3.0i+2.0j)"m/s"^(2). (a) What is the y-coordinates of the particle at the instant its x-coordinates is 84m? (b) what is the speed of the particle at this time?

A particle starts from origin at t= 0 with a velocity 5.0hat(i)m/s and

1.	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 (3.0i+2.0j)"m/s"^(2). (a) What is the y-coordinates of the particle at the instant its x-coordinates 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)at^(2)=5.0hat(i)t+"(1/2)"(3.0hat(i)+2.0hat(j))t^(2)` `=(5.0t+1.5t^(2))HAT(i)+1.0t^(2)hat(j)` Therefore, `x(t)=5.0t+1.5^(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+3.0t)hat(i)+2.0that(j)` At t = 6s, `v=23.0hat(i)+12.0hat(j)` SPEED `=\|v\|=sqrt(23^(2)+12^(2))=26ms^(-1)`