Two particles A and B move from rest along a | vedclass

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(C) Let the final velocity be $v$. For particle $B$,$v = f't$,so $t = \frac{v}{f'}$. The distance covered is $s = \frac{1}{2}f't^2 = \frac{1}{2}f'\lef

Vedclass · Accepted Answer

$\left(f^{\prime}-f\right) n=\frac{1}{2} f f^{\prime} m^{2}$

Vedclass · Answer

(C) Let the final velocity be $v$. For particle $B$,$v = f't$,so $t = \frac{v}{f'}$. The distance covered is $s = \frac{1}{2}f't^2 = \frac{1}{2}f'\left(\frac{v}{f'}\right)^2 = \frac{v^2}{2f'}$.For particle $A$,$v = f(t+m)$,so $t+m = \frac{v}{f}$,which means $t = \frac{v}{f} - m$. The distance covered is $s+n = \frac{1}{2}f(t+m)^2 = \frac{1}{2}f\left(\frac{v}{f}\right)^2 = \frac{v^2}{2f}$.From the velocity equations: $f't = f(t+m) \implies t(f'-f) = fm \implies t = \frac{fm}{f'-f}$.Substituting $t$ into the velocity equation $v = f't$: $v = \frac{f'fm}{f'-f}$.Now,$n = (s+n) - s = \frac{v^2}{2f} - \frac{v^2}{2f'} = \frac{v^2}{2} \left(\frac{f'-f}{ff'}\right)$.Substituting $v^2 = \left(\frac{ff'm}{f'-f}\right)^2$:$n = \frac{1}{2} \left(\frac{ff'm}{f'-f}\right)^2 \left(\frac{f'-f}{ff'}\right) = \frac{1}{2} \frac{(ff')^2 m^2}{(f'-f)^2} \cdot \frac{f'-f}{ff'} = \frac{ff'm^2}{2(f'-f)}$.Rearranging gives: $(f'-f)n = \frac{1}{2}ff'm^2$.

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

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