Question

If \( V_A \) and \( V_B \) are velocities of A and B respectively and their directions are normal to each other, the relative velocity of A with respect to B is

• A. \( V_A + V_B \)
• B. \( V_A - V_B \)
• C. \( \sqrt{V_A^2 + V_B^2} \)
• D. \( \sqrt{V_A^2 - V_B^2} \)

Hint:

When velocities are perpendicular to each other, the relative velocity is found using the Pythagorean theorem.

Answer 1

The Correct Option is C

Step 1: Use the formula for relative velocity.
When two velocities are perpendicular (normal to each other), the magnitude of the relative velocity \( V_{AB} \) of A with respect to B is given by: \[ V_{AB} = \sqrt{V_A^2 + V_B^2}. \]
Step 2: Conclusion.
Thus, the relative velocity of A with respect to B is \( \sqrt{V_A^2 + V_B^2} \), corresponding to option (C).

Answer 2

The relative velocity of A with respect to B is the vector \( \vec{V}_{AB} = \vec{V}_A - \vec{V}_B \). Since \( \vec{V}_A \) and \( \vec{V}_B \) are perpendicular, draw \( \vec{V}_A \) and \( -\vec{V}_B \) as two perpendicular sides of a right triangle; the resultant \( \vec{V}_{AB} \) is the hypotenuse. By the Pythagorean theorem, the magnitude of the hypotenuse is \( \sqrt{V_A^2 + V_B^2} \). Let's check each option against this vector-diagram result.

1. \( V_A + V_B \): This is the result of simply adding the magnitudes as if the vectors were parallel and pointing the same way, which ignores that they are perpendicular, not collinear.
2. \( V_A - V_B \): This treats the vectors as parallel and opposing, which is also incorrect since perpendicular vectors cannot be subtracted as plain magnitudes.
3. \( \sqrt{V_A^2 + V_B^2} \): This is exactly the hypotenuse length obtained from the right-triangle vector diagram, matching the perpendicular condition given.
4. \( \sqrt{V_A^2 - V_B^2} \): This would only apply if one vector were a component of the other along the same line, which is not the case for two mutually perpendicular vectors.

The right-triangle vector diagram confirms the magnitude of the relative velocity is \( \sqrt{V_A^2 + V_B^2} \).

Therefore, the correct answer is \( \sqrt{V_A^2 + V_B^2} \).