Question

Two particles $A$ and $B$ move from rest along a straight line with constant accelerations $f$ and $h$ respectiely. If $A$ takes $m$ seconds more than $B$ and describes $n$ units more than that of $B$ acquiring the same speed, then

• A. $(f+h)m^2=fhn$
• B. $(f-fh)m^2=fhn$
• C. $(h-f)n=\frac{1}{2}fh{m}^2$
• D. $\frac{1}{2}(f+h)n=fh{m}^2$

Answer 1

The correct option is C $(h-f)n=\frac{1}{2}fh{m}^2$

Using equation of motion $s=ut+\frac{1}{2}a{t}^2$

For $A:$
$S+n=\frac{1}{2}f(t+m{)}^2...(1)$ and
For $B:$
$S=\frac{1}{2}h{t}^2...\left(2\right)$

Using equation of motion $v=u+at$
For $A:v=f(t+m)...\left(3\right)$
For $B:v=ht...\left(4\right)$

From eqn$\left(3\right)$ and $\left(4\right)$
$f(t+m)=ht$
$\Rightarrow t(h-f)=fm$
$\Rightarrow t=\frac{fm}{h-f}$

From eqn$\left(1\right)$ and $\left(2\right)$
$S+n=\frac{1}{2}f{(\frac{fm}{h-f}+m)}^2...\left(5\right)$
$S=\frac{1}{2}h{\left(\frac{fm}{h-f}\right)}^2...\left(6\right)$

From eqn$\left(5\right)$ and $\left(6\right)$
$n=\frac{1}{2}f{(\frac{fm}{h-f}+m)}^2-\frac{1}{2}h{\left(\frac{fm}{h-f}\right)}^2$
$\Rightarrow n=\frac{1}{2}{m}^2[f{\left(\frac{h}{h-f}\right)}^2-h{\left(\frac{f}{h-f}\right)}^2]$
$\Rightarrow n=\frac{1}{2}{m}^2\frac{fh}{h-f}$
$\Rightarrow (h-f)n=\frac{1}{2}fh{m}^2$