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A river flows with a uniform velocity v. A person in a motorboat travels 0.855 km upstream, at which time a log is seen floating by. The person continues to travel upstream for 29.3 min at the same speed and then returns downstream to the starting point, where the same log is seen again. Find the flow velocity of the river. Assume the speed of the boat with respect to the water is constant throughout the entire trip. (Hint: The time of travel of the boat after it meets the log equals the time of travel of the log.)

Sagot :

Okpalawalter8idn

WE have that the flow velocity is

[tex]v=333.3 mts/h[/tex]

Distance traveled = 691 m

time = 38.1 min

speed(boat) =[tex]\frac{ 691}{38.1} =18.14 m/min[/tex]  

Distance traveled = 691 m

Generally the equation for down stream speed   is mathematically given as

[tex]t= 691/(18.14-v)[/tex]

Where

down stream speed = (18.14+v)

[tex]d = 2*691 m[/tex]

Therefor

[tex]t=\frac {2*691}{(18.14+v)}[/tex]

Generally the total time to reach starting point is

[tex]T= \frac{2*691}{(18.14+v)}+ \frac{691}{(18.14-v)}[/tex]

To find the speed of the log v is

d= 691 m

t= 691/v

[tex]\frac{691}{v }= \frac{2*691}{(18.14+v)}+ \frac{691}{(18.14-v)}\\\\\691(18.14^2-v^2)=691v(18.14+v)+1382v(18.14-v)\\\\227380-691v^2=12534.74V+691v^2+25069.48V-1382v^2[/tex]

Therefore

[tex]V=2.756*10^{-3} m/s\\\\v= \frac{10^9}{3} *10^6[/tex]

[tex]v=333.3 mts/h[/tex]

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