Q41 / 50 · Manipal (MET) 2024 · Physics

Two points move in the same straight line starting at the same moment from the same point in it. The first moves with constant velocity $u$ and the second with constant acceleration $f$. During the time elapses before the second catches, the first greatest distance between the particle is

  • A
    $\frac{u}{f}$
  • Correct answer
    B
    $\frac{u^2}{2 f}$
  • C
    $\frac{f}{2 u^2}$
  • D
    $\frac{f}{u^2}$

Solution

The greatest distance between the particles will be at the moment when velocity of both particles gets equal \(\therefore\) Using \(v=u+a t\), for acceleratod particle \(\begin{aligned} & u=0+(F)(t) \\ & t=\frac{u}{(F)} \end{aligned}\) Distance travelled by accelerated in time \(t\), \(\begin{aligned} D_1 & =O+\frac{1}{2}(F)\left(\frac{u}{F}\right)^2 \\ D_1 & =\frac{u^2}{2 F} \end{aligned}\) Distance travelled by constantly moving particle, \(\begin{aligned} & D_2=(u)\left(\frac{u}{F}\right)+\frac{1}{2}(0)\left(\frac{u}{F}\right)^2 \\ & D_2=\frac{u^2}{F} \end{aligned}\) Greatest distance between two particles \(\begin{gathered} D=\frac{u^2}{f}-\frac{u^2}{2 f} \\ D=\frac{u^2}{2 f} \end{gathered}\)

Hence, the correct answer is option: 'B'.

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