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

From Eq. (4.34a) for `r_(0)=0`, the position of the particle is given by
`r(t) = v_(0)t + 1/2at^(2)`
`=5.0hat(i)t + (1//2)(3.0hat(i)+2.0hat(j))t^(2)`
`=(5.0t + .5t^(2))hat(i) + 10t^(2)hat(j)`
Therefore, `x(t) = 5.0t + 1.5t^(2)`
y(t) = `+1.0t^(2)` ltrbgt Given, x(t) = 84m, t=?
`5.0t + 1.5t^(2)= 84 rArr t=6s`
At t=6s, y = 10 `(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))=~ 26 ms^(-1)`