COMEDK · Physics · 1. Mathematics in Physics
Two boys are standing at ends \(A\) and \(B\) of a ground where \(A B=200 \mathrm{~m}\). The boy at \(B\) starts running in a direction perpendicular to \(A B\) with a speed of \(6 \mathrm{~ms}^{-1}\). The boy at \(A\) starts simultaneously with a velocity of \(10 \mathrm{~ms}^{-1}\) and catches the other at time, \(t\) where the time, \(t\) is
- A \(50 \mathrm{~s}\)
- B \(20 \mathrm{~s}\)
- C \(25 \mathrm{~s}\)
- D \(12.5 \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(25 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below \[ \begin{aligned} &v_{A}=10 \mathrm{~m} / \mathrm{s} \\ &v_{B}=6 \mathrm{~m} / \mathrm{s} \end{aligned} \] The two boys meet at point \(C\) after a time \(t\). Horizontal component of velocity of \(v_{A}\).…
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