$A$ particle starts from the origin at $t=0$ with a velocity $5.0 \hat{i} \; m/s$ and moves in the $x-y$ plane under the action of a force which produces a constant acceleration of $(3.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
$(a)$ What is the $y$-coordinate of the particle at the instant its $x$-coordinate is $84 \; m$?
$(b)$ What is the speed of the particle at this time?

(A) The position vector of the particle at time $t$ is given by $r(t) = v_0 t + \frac{1}{2} a t^2$.
Given $v_0 = 5.0 \hat{i} \; m/s$ and $a = (3.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$,we have:
$r(t) = (5.0 \hat{i})t + \frac{1}{2}(3.0 \hat{i} + 2.0 \hat{j})t^2 = (5.0t + 1.5t^2) \hat{i} + (1.0t^2) \hat{j}$.
Thus,$x(t) = 5.0t + 1.5t^2$ and $y(t) = 1.0t^2$.
For $x = 84 \; m$,we solve $1.5t^2 + 5.0t - 84 = 0$. Using the quadratic formula,$t = \frac{-5.0 \pm \sqrt{25 + 4(1.5)(84)}}{2(1.5)} = \frac{-5.0 \pm \sqrt{529}}{3} = \frac{-5.0 \pm 23}{3}$. Since $t > 0$,$t = 6 \; s$.
$(a)$ At $t = 6 \; s$,$y = 1.0(6)^2 = 36.0 \; m$.
$(b)$ The velocity vector is $v(t) = \frac{dr}{dt} = (5.0 + 3.0t) \hat{i} + (2.0t) \hat{j}$.
At $t = 6 \; s$,$v = (5.0 + 3.0(6)) \hat{i} + (2.0(6)) \hat{j} = 23.0 \hat{i} + 12.0 \hat{j} \; m/s$.
The speed is $|v| = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx 25.94 \; m/s \approx 26 \; m/s$.