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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
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Verified Answer
The correct answer is:
$\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}$
$\begin{array}{llll}A:\quad u=0 & a_{1}=f & t_{1}=t+m & s_{1}=n+s \\ B: \quad u=0 & a_{2}=f^{\prime} & t_{2}=t & s_{2}=s\end{array}$
$\mathrm{s}+\mathrm{n}=\frac{1}{2} \mathrm{f} .(\mathrm{t}+\mathrm{m})^{2} \ldots \ldots(\mathrm{i}) \quad$ 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)
$\frac{1}{2} f^{\prime} \cdot\left(\frac{f m}{f^{\prime}-f}\right)^{2}+n=\frac{1}{2} f\left(\frac{f m}{f^{\prime}-f}+m\right)^{2}$
$\mathrm{s}+\mathrm{n}=\frac{1}{2} \mathrm{f} .(\mathrm{t}+\mathrm{m})^{2} \ldots \ldots(\mathrm{i}) \quad$ 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)
$\frac{1}{2} f^{\prime} \cdot\left(\frac{f m}{f^{\prime}-f}\right)^{2}+n=\frac{1}{2} f\left(\frac{f m}{f^{\prime}-f}+m\right)^{2}$
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