Question

A man row a boat at 4 kmhr^—1 in a still water , he row the boat 2km up stream and 2km back to it starting place In 2 hours ,how far is the stream flowing ?

Answer 1

Correct Question :

A man can row a boat at 4 km/hr in still water. He rows the boat 2 km upstream and 2 km back to his starting place in 2 hours. How fast is the stream flowing?

Answer:

The stream flows at 2√2 km/hr.

Explanation:

Given,

• A man can row a boat at 4 km/h in still water
• he rows the boat 2km upstream and 2km back to it starting place in 2 hours

To find,

• speed of the stream

Formula,

$\sf Speed =\dfrac{Distance}{time}\\\\ \underline{\boxed{\sf time=\dfrac{Distance}{Speed}}}$

Solution,

 Let Vₘ km/h be the speed of man in still water

and Vₛ km/h be the speed of the stream

t₁ - time taken to row 2 km upstream

t₂ - time taken to row 2 km downstream

• If a man rows a boat in the direction opposite to the flow of the stream, then it is called upstream.

 In this case,

net speed (upstream) = Speed of man in still water - speed of the stream

           = Vₘ - Vₛ

time taken to row 2 km upstream,

 $\sf t_1=\dfrac{2}{V_m-V_s}\\\\ t_1=\dfrac{2}{4 -V_s}$

• If the man rows a boat along the direction of the stream, then it is called downstream.

 In this case,

 net speed (downstream) = Speed of man in still water + speed of the stream

    = Vₘ + Vₛ

time taken to row 2 km downstream,

 $\sf t_2=\dfrac{2}{V_m+V_s}\\\\ t_2=\dfrac{2}{4+V_s}$

Also given,

total time taken to row 2 km upstream and 2 km downstream = 2 hours

  2 = t₁ + t₂

 $\sf 2=\dfrac{2}{4-V_s}+\dfrac{2}{4+V_s} \\\\ 2=2(\dfrac{1}{4-V_s}+\dfrac{1}{4+V_s} ) \\\\ 1=\dfrac{1}{4-V_s}+\dfrac{1}{4+V_s} \\\\ 1=\dfrac{4+V_s+4-V_s}{(4-V_s)(4+V_s)} \\\\ (4-V_s)(4+V_s)=8 \\\\ 4(4+V_s)-V_s(4+V_s)=8 \\\\ 16+4V_s-4V_s-V_s^2 =8 \\\\ 16-V_s^2=8 \\\\ V_s^2=16-8 \\\\ V_s^2=8 \\\\ V_s=2\sqrt{2} \ km/h$

∴ The stream flows at 2√2 km/hr.