Two points $A$ and $B$ move from rest along a straight line with constant acceleration $f$ and $f$ '…

Two points $A$ and $B$ move from rest along a straight line with constant acceleration $f$ and $f$ ' respectively. If $A$ takes $m$ sec. more than $B$ and describes ' $n$ ' units more than $B$ in acquiring the same speed then
  1. $\left(f-f^{\prime}\right) m^2=f f^{\prime} n$
  2. $\left(f+f^{\prime}\right) m^2=f f^{\prime} n$
  3. $\frac{1}{2}\left(f+f^{\prime}\right) m=f f^{\prime} n^2$
  4. $\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^2$

Solution

$ \begin{aligned} & v^2=2 f(d+n)=2 f^{\prime} d \\ & v=f^{\prime}(t)=(m+t) f \end{aligned} $ eliminate $d$ and $m$ we get $ \left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^2 \text {. } $

Asked in: JEE Main 2005

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