MHT CET · Physics · Motion In Two Dimensions
Two boys are standing at points \(A\) and \(B\) on ground, where distance \(A B=x\). The boy at \(B\) stars running perpendicular to \(A B\) with velocity \(V_1\). The boy at A starts running simultaneously with velocity v and meets the other boy in time \(t\). The value of \(t\) is
- A \(\left[\frac{x}{v-v_1}\right]^{1 / 2}\)
- B \(\left[\frac{x}{v_1-v}\right]^{1 / 2}\)
- C \(\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}\)
- D \(\left[\frac{x^2}{v_1^2-v^2}\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Let the coordinates of A be \((0,0)\) and B be \((x,0)\). Position of boy B at time \(t\): \((x, V_1 t)\).
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