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

Question

A body starts from rest and moves with constant acceleration. If it covers a distance of $ 40 \, \text{m} $ in 4 seconds, then its acceleration is:

Answer 1

Step 1: Set up axes so $ \vec{V}_A $ lies along the x-axis and $ \vec{V}_B $ lies along the y-axis, since the two are perpendicular: $ \vec{V}_A = (V_A, 0) $, $ \vec{V}_B = (0, V_B) $.
Step 2: The relative velocity vector is $ \vec{V}_{AB} = \vec{V}_A - \vec{V}_B = (V_A, -V_B) $.
Step 3: Its magnitude is $ |\vec{V}_{AB}| = \sqrt{V_A^2 + (-V_B)^2} = \sqrt{V_A^2 + V_B^2} $.
\[ \boxed{V_{AB} = \sqrt{V_A^2 + V_B^2}} \]

Answer 2

A more general way to find relative velocity magnitude uses the law of cosines: for two vectors with an angle $ \theta $ between them, $ V_{AB} = \sqrt{V_A^2 + V_B^2 - 2V_AV_B\cos\theta} $. Since $ \vec{V}_A $ and $ \vec{V}_B $ are normal (perpendicular) to each other, $ \theta = 90^\circ $ and $ \cos 90^\circ = 0 $, so the formula reduces to $ V_{AB} = \sqrt{V_A^2 + V_B^2} $. Let's check each option using this general formula.

$ V_A + V_B $: This would only be correct if $ \theta = 180^\circ $ (vectors directly opposed), where $ \cos\theta=-1$ makes the formula reduce to this sum — not the case here since $ \theta = 90^\circ $.
$ V_A - V_B $: This would only apply if $ \theta = 0^\circ $ (vectors in the same direction), where $ \cos\theta=1 $ — again not matching the given perpendicular condition.
$ \sqrt{V_A^2 + V_B^2} $: Substituting $ \theta = 90^\circ $ into the law-of-cosines formula gives exactly this expression, since the cross term vanishes when $ \cos\theta = 0 $.
$ \sqrt{V_A^2 - V_B^2} $: This expression does not arise from the law-of-cosines formula for any real angle $ \theta $ between two vectors added this way, so it cannot be correct.

The general law-of-cosines derivation, specialized to $ \theta = 90^\circ $, confirms the relative velocity magnitude is $ \sqrt{V_A^2 + V_B^2} $.

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

Answer 3

A more general way to find relative velocity magnitude uses the law of cosines: for two vectors with an angle $ \theta $ between them, $ V_{AB} = \sqrt{V_A^2 + V_B^2 - 2V_AV_B\cos\theta} $. Since $ \vec{V}_A $ and $ \vec{V}_B $ are normal (perpendicular) to each other, $ \theta = 90^\circ $ and $ \cos 90^\circ = 0 $, so the formula reduces to $ V_{AB} = \sqrt{V_A^2 + V_B^2} $. Let's check each option using this general formula.

$ V_A + V_B $: This would only be correct if $ \theta = 180^\circ $ (vectors directly opposed), where $ \cos\theta=-1$ makes the formula reduce to this sum — not the case here since $ \theta = 90^\circ $.
$ V_A - V_B $: This would only apply if $ \theta = 0^\circ $ (vectors in the same direction), where $ \cos\theta=1 $ — again not matching the given perpendicular condition.
$ \sqrt{V_A^2 + V_B^2} $: Substituting $ \theta = 90^\circ $ into the law-of-cosines formula gives exactly this expression, since the cross term vanishes when $ \cos\theta = 0 $.
$ \sqrt{V_A^2 - V_B^2} $: This expression does not arise from the law-of-cosines formula for any real angle $ \theta $ between two vectors added this way, so it cannot be correct.

The general law-of-cosines derivation, specialized to $ \theta = 90^\circ $, confirms the relative velocity magnitude is $ \sqrt{V_A^2 + V_B^2} $.

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

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

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The Correct Option is C

Approach Solution - 1

Approach Solution -2

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