A particle moves so that x = 3t² and y = 4t, in metres, with t in seconds…

Step 1. Velocity. $v_x = 6t$ and $v_y = 4$, so $\vec{v} = (6t\,\hat{i} + 4\hat{j})\ \text{m/s}$.

Step 2. Acceleration. $a_x = 6\ \text{m/s}^2$ and $a_y = 0\ \text{m/s}^2$, so $\vec{a} = 6\hat{i}\ \text{m/s}^2$, the same at every moment.

Step 3. Speed at 2 s. $\vec{v} = (12\hat{i} + 4\hat{j})\ \text{m/s}$, so the speed is $\sqrt{(12\ \text{m/s})^2 + (4\ \text{m/s})^2} = 12.6\ \text{m/s}$.

Step 4. At 45°. $6t = 4$ gives $t = 0.667\ \text{s}$.