Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $f$ ' respectively. If $A$ takes $\mathrm{m}$ sec. more than that of $\mathrm{B}$ and describes $\mathrm{n}$ units more than that of $\mathrm{B}$ in acquiring the same velocity, then

• A

  $\left(f+f^{\prime}\right) m^{2}=$ ff'n
• B

  $\left(f-f f^{\prime}\right) m^{2}=f f^{\prime} n$
• Correct answer
  C

  $\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}$
• D

  $\frac{1}{2}\left(\mathrm{f}+\mathrm{f}^{\prime}\right) \mathrm{m}=\mathrm{ff}^{\prime} \mathrm{n}^{2}$

Solution

$\begin{array}{llll}A:\quad u=0 & a_{1}=f & t_{1}=t+m & s_{1}=n+s \\ B: \quad u=0 & a_{2}=f^{\prime} & t_{2}=t & s_{2}=s\end{array}$ $\mathrm{s}+\mathrm{n}=\frac{1}{2} \mathrm{f} .(\mathrm{t}+\mathrm{m})^{2} \ldots \ldots(\mathrm{i}) \quad$ and $\mathrm{s}=\frac{1}{2} \mathrm{f}^{\prime}(\mathrm{t})^{2} \ldots(\mathrm{ii}) \quad \therefore \mathrm{f}^{\prime}(\mathrm{t})^{2}+\mathrm{n}=\frac{1}{2} \mathrm{f}(\mathrm{t}+\mathrm{m})^{2} \ldots (iii)$ $\begin{array}{l} v_{1}=u_{1}+a_{1} t_{1}=0+f .(t+m) \\ v_{2}=u_{2}+a_{2} t_{2}=0+f^{\prime} . t \\ \therefore f(t+m)=f^{\prime} t \\ t=\frac{f m}{f^{\prime}-f} \end{array}$ from (iii) $\frac{1}{2} f^{\prime} \cdot\left(\frac{f m}{f^{\prime}-f}\right)^{2}+n=\frac{1}{2} f\left(\frac{f m}{f^{\prime}-f}+m\right)^{2}$

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