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JIPMER 2013
Exam
JIPMER
Year
2013
Subject
Physics
Chapter
Motion in One Dimension

JIPMER 2013 · Physics · Motion in One Dimension

Two boys are standing at the ends A and B of a ground, where A B=a...

Two boys are standing at the ends AA and BB of a ground, where AB=aA B=a. The boy at BB starts running in a direction perpendicular to ABA B with velocity v1v_1. The boy at AA starts running simultaneously with velocity vv and catches the other boy in a time tt, where tt is

Options

  1. Aav2+v12\frac{a}{\sqrt{v^2+v_1^2}}
  2. Ba2v2−v12\sqrt{\frac{a^2}{v^2-v_1^2}}
  3. C\frac{a}{\left(v-v_1\right)}
  4. D\frac{a}{\left(v+v_1\right)}

Correct answer

B. a2v2−v12\sqrt{\frac{a^2}{v^2-v_1^2}}

Step-by-step solution

Distance covered by boy Aintimein timet A C=v t\ldots(i)DistancecoveredbyboyDistance covered by boyBintimein timet B C=v_1 t\ldots(ii)UsingPythagorastheoremUsing Pythagoras theoremAC2=AB2+BC2 or  (vt)2=a2+(v1t)2\begin{aligned} A C^2 & =A B^2+B C^2 \\ \text{\text{ or } } \quad(v t)^2 & =a^2+\left(v_1 t\right)^2\end{aligned}orv2t2−v12t2=a2ort2or \quad v^2 t^2-v_1^2 t^2=a^2 or \quad t^2\left(v^2-v_1^2\right)=a^2 \therefore t=\sqrt{\frac{a^2}{\left(v^2-v_1^2\right)}}
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