Find Velocity Vector: Velocity $\vec{v}(t)$ is the time derivative of the position vector $\vec{r}(t)$. Given $\vec{r}(t) = 4t\hat{i} + 2t^2\hat{j} + 5\hat{k}$. $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(4t)\hat{i} + \frac{d}{dt}(2t^2)\hat{j} + \frac{d}{dt}(5)\hat{k}$ $\vec{v}(t) = 4\hat{i} + 4t\hat{j}$

Calculate Velocity at t = 1 s: Substitute $t = 1$ into the velocity equation: $\vec{v}(1) = 4\hat{i} + 4(1)\hat{j} = 4\hat{i} + 4\hat{j} \text{ ms}^{-1}$

Calculate Magnitude: The magnitude of the velocity vector is: $|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 4^2}$ $|\vec{v}| = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \text{ ms}^{-1}$

Calculate Direction: The angle $\theta$ with the x-axis is given by: $\tan \theta = \frac{v_y}{v_x} = \frac{4}{4} = 1$ $\theta = \tan^{-1}(1) = 45^{\circ}$ Thus, the magnitude is $4\sqrt{2} \text{ ms}^{-1}$ and the direction is $45^{\circ}$ with the x-axis.