Question

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 \(t = 1\) s the angle made by the velocity vector with x-axis is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $120^\circ$
• E. $90^\circ$

Hint:

For vector problems: - Velocity = derivative of position vector - Use $\tan\theta = \frac{v_y}{v_x}$ for angle with x-axis

Answer 1

The Correct Option is C

Concept: Velocity is the derivative of position vector: \[ \vec{v} = \frac{d\vec{r}}{dt} \] Angle with x-axis is given by: \[ \tan \theta = \frac{v_y}{v_x} \]

Step 1: Differentiate position vector.
\[ \vec{r} = 2t\hat{i} + \sqrt{3}t^2\hat{j} + 5\hat{k} \] \[ \vec{v} = \frac{d\vec{r}}{dt} = 2\hat{i} + 2\sqrt{3}t\hat{j} \]

Step 2: Substitute $t = 1$.
\[ \vec{v} = 2\hat{i} + 2\sqrt{3}\hat{j} \]

Step 3: Find angle with x-axis.
\[ \tan \theta = \frac{2\sqrt{3}}{2} = \sqrt{3} \]

Step 4: Determine $\theta$.
\[ \theta = 60^\circ \]