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 \(t = 1\,s\):

• A. \(\sqrt{10} \, m\,s^{-1}\)
• B. \(10 \, m\,s^{-1}\)
• C. \(\sqrt{3} \, m\,s^{-1}\)
• D. \(3 \, m\,s^{-1}\)

Hint:

Relative velocity is always found by vector subtraction: \( \vec{v}_{AB} = \vec{v}_A - \vec{v}_B \).

Answer 1

The Correct Option is A

Step 1: Velocity of particle A.
Moves along x-axis: v_A = (1, 0) m s⁻¹.

Step 2: Velocity of particle B.
Given y = t³, v_B = dy/dt = 3t². At t = 1 s, v_B = (0, 3) m s⁻¹.

Step 3: Relative velocity calculation.
v_AB = v_A - v_B = (1, -3).

Step 4: Magnitude determination.
|v_AB| = √(1² + (-3)²) = √(1+9) = √10.

Step 5: Conclusion.
Hence, the relative speed is √10 m s⁻¹.