$A$ particle starts from the origin at $t=0 \; s$ with a velocity of $10.0 \hat{j} \; m/s$ and moves in the $x-y$ plane with a constant acceleration of $(8.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
$(a)$ At what time is the $x$-coordinate of the particle $16 \; m$? What is the $y$-coordinate of the particle at that time?
$(b)$ What is the speed of the particle at that time?

(A) Given: Initial velocity $\vec{u} = 10.0 \hat{j} \; m/s$,Acceleration $\vec{a} = (8.0 \hat{i} + 2.0 \hat{j}) \; m/s^2$.
Using the kinematic equation for position: $\vec{r}(t) = \vec{u}t + \frac{1}{2} \vec{a}t^2$.
Substituting the values: $\vec{r}(t) = (10.0 \hat{j})t + \frac{1}{2} (8.0 \hat{i} + 2.0 \hat{j})t^2 = 4.0 t^2 \hat{i} + (10t + t^2) \hat{j}$.
Equating components: $x = 4.0 t^2$ and $y = 10t + t^2$.
For $x = 16 \; m$: $16 = 4.0 t^2 \implies t^2 = 4 \implies t = 2 \; s$.
At $t = 2 \; s$,$y = 10(2) + (2)^2 = 20 + 4 = 24 \; m$.
$(b)$ Velocity vector $\vec{v}(t) = \vec{u} + \vec{a}t = 10.0 \hat{j} + (8.0 \hat{i} + 2.0 \hat{j})t = 8.0 t \hat{i} + (10 + 2.0 t) \hat{j}$.
At $t = 2 \; s$: $\vec{v} = 8.0(2) \hat{i} + (10 + 2.0(2)) \hat{j} = 16 \hat{i} + 14 \hat{j}$.
Speed $|\vec{v}| = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} \approx 21.26 \; m/s$.