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Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $h,$ respectively. If $A$ takes $m$ seconds more than $B$ and describes $n$ units more than that of $B$ acquiring the same speed, then
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The correct answers are:
$(h-f) n=\frac{1}{2}$ $fhm^{2}$
Let $B$ travels $x$ units, $v=u+a t$
According to problem, $h t=f(t+m)$ $\begin{aligned} h t=& f t+f m \\ h t-f t &=f m \\ t(h-f) &=f m \\ \frac{h-f}{f} &=\frac{m}{t} \end{aligned}$
$t=m\left(\frac{f}{h-f}\right)$
$\Rightarrow \quad t^{2}=m^{2}\left(\frac{f}{h-f}\right)^{2}$
Again, $n+x=\frac{1}{2} f(t+m)^{2}$
$\Rightarrow \quad h+\frac{1}{2} h t^{2}=\frac{1}{2} f(t+m)^{2}$
From Eqs. (i) and (ii) $\frac{m}{n}=\frac{2(h-f)}{m f h}$
$\Rightarrow \quad(h-f) n=\frac{m^{2} f h}{2}$
According to problem, $h t=f(t+m)$ $\begin{aligned} h t=& f t+f m \\ h t-f t &=f m \\ t(h-f) &=f m \\ \frac{h-f}{f} &=\frac{m}{t} \end{aligned}$
$t=m\left(\frac{f}{h-f}\right)$
$\Rightarrow \quad t^{2}=m^{2}\left(\frac{f}{h-f}\right)^{2}$
Again, $n+x=\frac{1}{2} f(t+m)^{2}$
$\Rightarrow \quad h+\frac{1}{2} h t^{2}=\frac{1}{2} f(t+m)^{2}$
From Eqs. (i) and (ii) $\frac{m}{n}=\frac{2(h-f)}{m f h}$
$\Rightarrow \quad(h-f) n=\frac{m^{2} f h}{2}$
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