A motorboat covers a given distance in 6 ,h m | vedclass

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(B) Let $v_{b}$ be the velocity of the motorboat in still water and $v_{w}$ be the velocity of the river water.When moving downstream,the effective ve

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$7.5$

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(B) Let $v_{b}$ be the velocity of the motorboat in still water and $v_{w}$ be the velocity of the river water.When moving downstream,the effective velocity is $(v_{b} + v_{w})$. The distance $x$ covered in $6 \,h$ is:$x = (v_{b} + v_{w}) 	imes 6$  ---$(i)$When moving upstream,the effective velocity is $(v_{b} - v_{w})$. The distance $x$ covered in $10 \,h$ is:$x = (v_{b} - v_{w}) 	imes 10$  ---(ii)Equating the two expressions for distance $x$:$(v_{b} + v_{w}) 	imes 6 = (v_{b} - v_{w}) 	imes 10$$6v_{b} + 6v_{w} = 10v_{b} - 10v_{w}$$16v_{w} = 4v_{b}$$v_{w} = \frac{v_{b}}{4}$Substituting $v_{w}$ back into equation $(i)$ to find $x$ in terms of $v_{b}$:$x = (v_{b} + \frac{v_{b}}{4}) 	imes 6 = (\frac{5v_{b}}{4}) 	imes 6 = 7.5v_{b}$The time $t$ taken to cover distance $x$ in still water (where velocity is $v_{b}$) is:$t = \frac{x}{v_{b}} = \frac{7.5v_{b}}{v_{b}} = 7.5 \,h$

$A$ motorboat covers a given distance in $6 \,h$ moving downstream on a river. It covers the same distance in $10 \,h$ moving upstream. The time it takes to cover the same distance in still water is (in $\,h$)

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