MHT CET · Physics · Motion In Two Dimensions
Two girls are standing at the ends 'A' and 'B' of a ground where \(A B=b\). The girl at ' \(B\) ' starts running in a direction perpendicular to ' AB ' with velocity ' \(\mathrm{V}_1\) '. The girl at 'A' starts running simultaneously with velocity ' \(\mathrm{V}_2\) ' and in shortest distance meets the other girl in time ' t '. The value of ' \(t\) ' is
- A \(\frac{b}{\sqrt{V_1{ }^2+V_2{ }^2}}\)
- B \(\frac{b}{V_1+V_2}\)
- C \(\frac{b}{V_2-V_1}\)
- D \(\frac{b}{\sqrt{V_2{ }^2-V_1{ }^2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{b}{\sqrt{V_2{ }^2-V_1{ }^2}}\)
Step-by-step Solution
Detailed explanation
\( (V_2 t)^2 = b^2 + (V_1 t)^2 \) \( V_2^2 t^2 - V_1^2 t^2 = b^2 \)
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