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 positon of the parricle at time t is given by
`vecr(t) = vecut +(1)/(2)vecat^(2)`
`" "=(5.0hati)t+(1)/(3)(3.0hati + 2.0 hatj)t^(2)`
`" "=(5.0hati + 1.5t^(2))hati + ( 1.0 t^(2))hatj" "......(i)`
`therefore" "x(t) = 5.0 t + 1.5 t^(2)`
`" "y(t) = 1.0 t^(2)`
Given that, `x(t) = 84 m, t = ?`
`therefore" " 84 = 5.0 t + 1.5 t^(2)`
Solving, `t = 6s`
At `" "t= 6s , y = 1.0 (6)^(2) = 36 m`
Differentiating Eqn. (i) w.r.t. time,
`" " vecV = (dvecr)/(dt) = (5.0 + 3.0t)hati + (2.0t)hati`
At `t = 6s, vecV = 23.0 hati + 12 hatj`
Speed `= |vecv| = sqrt((23)^(2) + (12)^(2))= 25.9 m s^(-1)`