Discussion
If the position vector of a particle is \(\vec{r} = 2t\hat{i} + \sqrt{3}t^{2}\hat{j} + 5\hat{k}\) with \(\vec{r}\) in m and t in s, then at t=1s the angle made by the velocity vector with x-axis is
Answer: C
The angle is 60°.
Differentiate the position vector to get the velocity:
\[ \begin{aligned} \vec{v} &= \frac{d\vec{r}}{dt} = 2\hat{i} + 2\sqrt{3}t\hat{j} \\ \vec{v}(1) &= 2\hat{i} + 2\sqrt{3}\hat{j} \end{aligned} \]
The velocity has no z-component, so it lies in the x-y plane, and its angle with the x-axis satisfies \(\tan\theta = \frac{2\sqrt{3}}{2} = \sqrt{3}\), giving \(\theta = 60^\circ\).
30° is the angle with the y-axis. Using the position vector instead of the velocity would bring in the constant 5 in the z-direction, which does not affect the velocity.
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