Q107 / 155 · WBJEE 2019 · Mathematics

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

  • A
    $(f+h) m^{2}=h n$
  • B
    $(f-f h) m^{2}=f h n$
  • Correct answer
    C
    $(h-f) n=\frac{1}{2}$ $fhm^{2}$
  • D
    $\frac{1}{2}(f+h) n=$ $fhm^{2}$

Solution

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) $\frac{m}{n}=\frac{2(h-f)}{m f h}$ $\Rightarrow \quad(h-f) n=\frac{m^{2} f h}{2}$

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

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