Question:
Particle A (at origin at \(t=0\)) moves along x-axis with velocity \(1 \, m\,s^{-1}\). Particle B moves along y-axis such that \(y = t^3\) (with \(c=1 \, m\,s^{-3}\)). Find the magnitude of relative velocity of A with respect to B at \(t = 1\,s\):
Particle A (at origin at \(t=0\)) moves along x-axis with velocity \(1 \, m\,s^{-1}\). Particle B moves along y-axis such that \(y = t^3\) (with \(c=1 \, m\,s^{-3}\)). Find the magnitude of relative velocity of A with respect to B at \(t = 1\,s\):
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Relative velocity is always found by vector subtraction: \( \vec{v}_{AB} = \vec{v}_A - \vec{v}_B \).
Updated On: Jun 19, 2026
- \(\sqrt{10} \, m\,s^{-1}\)
- \(10 \, m\,s^{-1}\)
- \(\sqrt{3} \, m\,s^{-1}\)
- \(3 \, m\,s^{-1}\)
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The Correct Option is A
Solution and Explanation
Step 1: Velocity of particle A.
Particle A moves along x-axis with constant velocity: \[ \vec{v}_A = (1,0) \]
Step 2: Position of particle B.
Given: \[ y = t^3 \] So velocity of B is: \[ v_B = \frac{dy}{dt} = 3t^2 \]
Step 3: Velocity of particle B at \(t=1\).
At \(t=1\): \[ \vec{v}_B = (0,3) \]
Step 4: Relative velocity of A with respect to B.
\[ \vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (1,0) - (0,3) = (1,-3) \]
Step 5: Magnitude of relative velocity.
\[ |\vec{v}_{AB}| = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10} \]
Step 6: Final conclusion.
Therefore, the magnitude of relative velocity is \( \sqrt{10} \, m\,s^{-1} \).
Final Answer: \[ \boxed{\sqrt{10} \, m\,s^{-1}} \]
Particle A moves along x-axis with constant velocity: \[ \vec{v}_A = (1,0) \]
Step 2: Position of particle B.
Given: \[ y = t^3 \] So velocity of B is: \[ v_B = \frac{dy}{dt} = 3t^2 \]
Step 3: Velocity of particle B at \(t=1\).
At \(t=1\): \[ \vec{v}_B = (0,3) \]
Step 4: Relative velocity of A with respect to B.
\[ \vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (1,0) - (0,3) = (1,-3) \]
Step 5: Magnitude of relative velocity.
\[ |\vec{v}_{AB}| = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10} \]
Step 6: Final conclusion.
Therefore, the magnitude of relative velocity is \( \sqrt{10} \, m\,s^{-1} \).
Final Answer: \[ \boxed{\sqrt{10} \, m\,s^{-1}} \]
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