A boatman goes 2, km against the current of t | vedclass

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Question

(B) Let the speed of the boat in still water be $u\, km/h$ and the speed of the stream be $v\, km/h$.Upstream speed $= u - v = \frac{2\, km}{1\, h} =

Vedclass · Accepted Answer

$1\, hour\, 15\, minutes$

Vedclass · Answer

(B) Let the speed of the boat in still water be $u\, km/h$ and the speed of the stream be $v\, km/h$.Upstream speed $= u - v = \frac{2\, km}{1\, h} = 2\, km/h$.Downstream speed $= u + v = \frac{1\, km}{10\, min} = \frac{1\, km}{1/6\, h} = 6\, km/h$.Adding the two equations: $(u - v) + (u + v) = 2 + 6 \implies 2u = 8 \implies u = 4\, km/h$.Thus, the speed of the boat in still water is $4\, km/h$.Time required to cover $5\, km$ in stationary water $= \frac{	ext{Distance}}{	ext{Speed}} = \frac{5}{4}\, hours$.$\frac{5}{4}\, hours = 1\, hour\, 15\, minutes$.

$A$ boatman goes $2\, km$ against the current of the stream in $1\, h$ and goes $1\, km$ along the current in $10\, min.$ How long will he take to go $5\, km$ in stationary water?

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