Question

Car P is heading east with a speed V and car Q is heading north with a speed \(\sqrt{3}\). What is the velocity of car Q with respect to car P?

• A. V\(\sqrt{3}\),heading north
• B. 2V,30° east of north
• C. V\(\sqrt{3}\),60° west of north
• D. 2V,30° west of north
• E. V\(\sqrt{2}\),45° west of north

Answer 1

The Correct Option is D

Given:

• Car \( P \) moves east with speed \( V \).
• Car \( Q \) moves north with speed \( \sqrt{3}V \).

Step 1: Represent Velocities as Vectors

Let the east direction be the positive \( x \)-axis and the north direction be the positive \( y \)-axis.

Velocity of \( P \): \( \overrightarrow{V_P} = V \hat{i} \).

Velocity of \( Q \): \( \overrightarrow{V_Q} = \sqrt{3}V \hat{j} \).

Step 2: Find Relative Velocity

The velocity of \( Q \) with respect to \( P \) is:

\[ \overrightarrow{V_{Q/P}} = \overrightarrow{V_Q} - \overrightarrow{V_P} = \sqrt{3}V \hat{j} - V \hat{i} \]

Step 3: Calculate Magnitude and Direction

Magnitude of \( \overrightarrow{V_{Q/P}} \):

\[ |\overrightarrow{V_{Q/P}}| = \sqrt{(-V)^2 + (\sqrt{3}V)^2} = \sqrt{V^2 + 3V^2} = 2V \]

Direction (angle \( \theta \) west of north):

\[ \tan \theta = \frac{V}{\sqrt{3}V} = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ \]

Thus, the direction is \( 30^\circ \) west of north.

Conclusion:

The velocity of car \( Q \) with respect to car \( P \) is \( 2V \), \( 30^\circ \) west of north.

Answer: \(\boxed{D}\)

Answer 2

Let $\vec{v_P}$ be the velocity of car P and $\vec{v_Q}$ be the velocity of car Q. 

$\vec{v_P} = V \hat{i}$ (east) $\vec{v_Q} = \sqrt{3}V \hat{j}$ (north) 

We want to find the velocity of car Q with respect to car P, which is denoted as $\vec{v_{QP}}$. $\vec{v_{QP}} = \vec{v_Q} - \vec{v_P} = \sqrt{3}V \hat{j} - V \hat{i}$ 

The magnitude of $\vec{v_{QP}}$ is given by: $|\vec{v_{QP}}| = \sqrt{(-V)^2 + (\sqrt{3}V)^2} = \sqrt{V^2 + 3V^2} = \sqrt{4V^2} = 2V$ 

The direction of $\vec{v_{QP}}$ can be found using the tangent function: $\tan{\theta} = \frac{\sqrt{3}V}{-V} = -\sqrt{3}$ $\theta = \arctan{(-\sqrt{3})} = -60°$ 

Since we have $-V$ in the x-direction and $\sqrt{3}V$ in the y-direction, the angle is in the second quadrant. 

The angle $-60°$ is equivalent to $180° - 60° = 120°$ measured counterclockwise from the positive x-axis. 

This angle represents 30° west of north. 

Thus, the velocity of car Q with respect to car P is 2V, 30° west of north. 

Final Answer: The final answer is $\boxed{D}$