91Ó°ÊÓ

The position vector is given by: \(\vec{r}=3 \hat{i}+ \sqrt{3} t^{2}\hat{j}-4 \hat{k}\). To find the velocity vector \(\vec{v}\), we need to differentiate \(\vec{r}\) with respect to time \(t\): \(\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3\hat{i}+\sqrt{3}t^2\hat{j}-4\hat{k})\) Now, differentiate each component of the vector with respect to \(t\): \(\vec{v} = (0\hat{i} + 2\sqrt{3}t\hat{j} + 0\hat{k})\)

Now, we need to find the velocity vector at \(t=\sqrt{3}\) seconds. Plug the value of \(t=\sqrt{3}\) into the velocity vector equation: \(\vec{v}(\sqrt{3}) = (0\hat{i} + 2\sqrt{3}{(\sqrt{3})}\hat{j} + 0\hat{k}) = 6\hat{j}\) At \(t=\sqrt{3}\) seconds, the velocity vector is \(\vec{v} = 6\hat{j}\).

To determine the magnitude of the velocity vector, use the following formula: Magnitude \(|v| = \sqrt{v_x^2+v_y^2+v_z^2}\) In our case, \(\vec{v}= 6\hat{j}\), so we have \(v_x=0\), \(v_y=6\), and \(v_z=0\). Plug in the values: \( |v| = \sqrt{(0)^2 + (6)^2 + (0)^2} = \sqrt{36} = 6 \mathrm{~m} / \mathrm{s}\) The magnitude of the velocity vector at \(t=\sqrt{3}\) s is \(6 \mathrm{~m} / \mathrm{s}\).

The velocity vector is purely in the vertically upward direction (i.e., along the \(y\)-axis), as there are no components in the \(x\) and \(z\)-directions. So, the angle with the horizontal \(x\)-axis (often denoted as \(\theta\)) is: \(\theta = \tan^{-1}\left(\frac{vy}{vx}\right) = \tan^{-1}\left(\frac{6}{0}\right)\) Since, the division by zero is undefined, it indicates that the vertical velocity is perpendicular to the horizontal direction. In this case, the angle will be: \(\theta = 90^{\circ}\) The direction angle of the velocity vector at \(t=\sqrt{3}\) seconds is \(90^{\circ}\). However, none of the answer choices match the solution we found. The provided answers may be incorrect, or there might be a mistake in the given position vector. In any case, the step-by-step approach outlined above is what you should follow when faced with similar problems.