Asked by chris
In a canoe race, a team paddles downstream 480 m. in 60 s. The same team travels up stream in 80 s. Find the team's rate in still water and the rate of the current.
Please show how the equation is set up to solve
Please show how the equation is set up to solve
Answers
Answered by
drwls
Let the distance paddled (one way) be L = 480 m
Let V be the canoe speed in still water and v be the stream speed.
L/(V+v) = 60 s
L/(V-v) = 80 s
(V+v)/(V-v) = 80/60 = 4/3
V + v = (4/3)V - (4/3)v
V/3 = (7/3)v
v = V/7
480/(8V/7) = 60
V = 7 m/s
v = 1 m/s
Let V be the canoe speed in still water and v be the stream speed.
L/(V+v) = 60 s
L/(V-v) = 80 s
(V+v)/(V-v) = 80/60 = 4/3
V + v = (4/3)V - (4/3)v
V/3 = (7/3)v
v = V/7
480/(8V/7) = 60
V = 7 m/s
v = 1 m/s
Answered by
caroline cropper
love math ice take
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