Question

Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $f^{\prime }$ 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. $\left(f+f^{\prime }\right)m^2=f{f}^{\prime }n$
• B. $\left(f-f{f}^{\prime }\right)m^2=f{f}^{\prime }n$
• C. $\left(f^{\prime }-f\right)n=\frac{1}{2}f{f}^{\prime }{m}^2$
• D. $\frac{1}{2}\left(f+f^{\prime }\right)m=f{f}^{\prime }{n}^2$

Answer 1

Answer: (C)

$A:u=0$
$a_1=f$
$t_1=t+m$
$s=n+s$
$B:u=0$
$a_2=f^{\prime }$
$t_2=t$
$s=s$
$s+n=\frac{1}{2}f.(t+m{)}^2\dots \dots$(i)
and $s=\frac{1}{2}{f}^{\prime }(t{)}^2\dots$ (ii)
$\therefore f^{\prime }(t{)}^2+n=\frac{1}{2}f(t+m{)}^2\dots \dots$ (iii)
$v_1=u_1+a_1{t}_1=0+f.(t+m)$
$v_2=u_2+a_2{t}_2=0+f^{\prime }.t$
$\therefore f(t+m)=f^{\prime }t$
$t=\frac{fm}{f^{\prime }-f}$
from (iii)
$\frac{1}{2}{f}^{\prime }\cdot {\left(\frac{fm}{f^{\prime }-f}\right)}^2+n$
$=\frac{1}{2}f{\left(\frac{fm}{f^{\prime }-f}+m\right)}^2$
$\left(f^{\prime }-f\right)n=\frac{1}{2}f{f}^{\prime }{m}^2$