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while canoeing ont he tickfaw river, shannon traveled 30 miles upstream against the current in 5 hr. It took her only 3 hr paddlingd ownstream with the current back to the spot where she started. Find shannons coeing rate in still water and the rate of the current

Respuesta :

Pot8o

Answer:

The canoeing speed is 8mi/h and the stream speed is 2mi/h.

Explanation:

Being Vc the speed of canoing and Vs the speed of the stream wmi/e have that

[tex]\left \{ {{ Vc - Vs = 30mi / 5hr } \atop {Vc + Vs = 30mi /3hs}} \right.[/tex]

then

[tex]\left \{ {{ Vc - Vs = 6mi/h } \atop {Vc + Vs = 10mi/h}} \right. [/tex]

using the first equation of the system we have that

Vc  = 6mi/h +Vs

replacing

6mi/h+Vs+Vs = 10mi/h

2Vs = 4mi/h

Vs = 2mi/h

then going back

Vc = 6mi/h + 2mi/h = 8 mi/h

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