Question

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

• A. $\left(f+f'\right) m^{2}=f f' n$
• B. $\left(f-f f'\right) m^{2}=f f' n$
• C. $\left(f'-f\right) n=\frac{1}{2} f f' m^{2}$
• D. $\frac{1}{2}\left(f+f'\right) m=f f'n^{2}$

Answer 1

Answer: (C)

$A: u=0 $
$a_{1}=f $
$t_{1}=t+m $
$s=n+s$
$B: u=0 $
$a_{2}=f' $
$ t_{2}=t $
$s=s$
$s+n=\frac{1}{2} f .(t+m)^{2} \ldots \ldots$(i)
and $s=\frac{1}{2} f^{\prime}(t)^{2} \ldots $ (ii)
$\therefore f'(t)^{2}+n=\frac{1}{2} f(t+m)^{2} \ldots \ldots $ (iii)
$v_{1}=u_{1}+a_{1} t_{1}=0+f .(t+m)$
$v_{2}=u_{2}+a_{2} t_{2}=0+f' . t$
$\therefore f(t+m)=f't$
$t=\frac{f m}{f'-f}$
from (iii)
$\frac{1}{2} f' \cdot\left(\frac{f m}{f'-f}\right)^{2}+n$
$=\frac{1}{2} f\left(\frac{f m}{f'-f}+m\right)^{2}$
$\left(f'-f\right) n=\frac{1}{2} f f'm^{2}$