A man is $ 45 \mathrm{~m} $ behind the bus when the bus start accelerating from rest with acceleration 2.5 $ m / s^{2} $. With what minimum velocity should the man start running to catch the bus.
(a) $ 12 \mathrm{~m} / \mathrm{s} $
(b) $ 14 \mathrm{~m} / \mathrm{s} $
(c) $ 15 \mathrm{~m} / \mathrm{s} $
(d) $ 16 \mathrm{~m} / \mathrm{s} $

$Given$

$Initial\ Velocity,( u) =0$

$a=2.5m/s^{2}$

$Distance\ between\ man\ and\ bus=45m$

$From\ 2nd\ equation\ of\ motion,$

$s=ut+\frac{1}{2} at^{2}$

$substituting\ the\ given\ value\ on\ the\ equation,\ we\ get\ distance\ travelled\ by\ the\ bus$

$s=0+\frac{1}{2} 2.5t^{2}$

$s=\frac{2.5t^{2}}{2} \ ...........................( 1)$

$Lets\ consider,\ u=initial\ velocity\ of\ the\ man$

$ \begin{array}{l} According\ to\ the\ question,\ man\ need\ to\ cover\ the\ distance\ ( 45+s) m\ at\ time\ t\ to\ catch\ the\ bus,\ \ it\ can\ be\ given\ as \end{array}$

$ut=45+s$

$ut=45+\frac{2.5t^{2}}{2}$

$ut=45+1.25t^{2}$

$u=\frac{45+1.25t^{2}}{t}$

$u=\frac{45}{t} +\frac{1.25t^{2}}{t}$

$u=\frac{45}{t} +1.25t\ ..........................( 2)$

$For\ minimum\ velocity,\ the\ minimum\ value\ of\ u\ will\ be:$

$\frac{du}{dt} =0$

$-\frac{45}{t^{2}} +1.25=0$

$-\frac{45}{t^{2}} =-1.25$

$45=1.25t^{2}$

$t^{2} =\frac{45}{1.25}$

$t^{2} =36$

$t=\sqrt{36}$

$t=6sec$

$From\ equation\ ( 2) ,$

$u_{min} =\frac{45}{6} +1.25\times 6$

$u_{min} =7.5+7.5$

$u_{min} =15m/s$

$Therefore,\ the\ minimum\ velocity\ that\ a\ man\ would\ need\ to\ start\ running\ to\ catch\ a\ bus\ is\ 15m/s.$

$So,\ the\ correct\ option\ will\ be\ ( c) .$

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Updated on: 10-Oct-2022

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