Two boys,A and B,run at 4.5 text{ km/h} and 6 | vedclass

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(A) First,convert the speeds from $	ext{km/h}$ to $	ext{m/s}$ by multiplying by $\frac{5}{18}$.Speed of $A = 4.5 	imes \frac{5}{18} = \frac{9}{2} \

Vedclass · Accepted Answer

$60$

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(A) First,convert the speeds from $	ext{km/h}$ to $	ext{m/s}$ by multiplying by $\frac{5}{18}$.Speed of $A = 4.5 	imes \frac{5}{18} = \frac{9}{2} 	imes \frac{5}{18} = 1.25 	ext{ m/s} = \frac{5}{4} 	ext{ m/s}$.Speed of $B = 6 	imes \frac{5}{18} = \frac{5}{3} 	ext{ m/s}$.$A$ has a head start of $190 	ext{ m}$,so $A$ only needs to cover $1000 - 190 = 810 	ext{ m}$ to reach the finish line.$B$ needs to cover the full $1000 	ext{ m}$.Time taken by $B$ to finish the race $= \frac{	ext{Distance}}{	ext{Speed}} = \frac{1000}{5/3} = 1000 	imes \frac{3}{5} = 600 	ext{ seconds}$.In these $600 	ext{ seconds}$,the distance covered by $A = 	ext{Speed} 	imes 	ext{Time} = \frac{5}{4} 	imes 600 = 750 	ext{ m}$.Since $A$ was supposed to cover $810 	ext{ m}$ to finish,but only covered $750 	ext{ m}$,the distance by which $B$ wins is $810 - 750 = 60 	ext{ m}$.

Two boys,$A$ and $B$,run at $4.5 \text{ km/h}$ and $6 \text{ km/h}$ respectively. $A$ is given a head start of $190 \text{ m}$. If the total course length is $1 \text{ km}$,by what distance (in $\text{m}$) does $B$ win?

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