Asked by Sydney
Question: The swim portion of a triathlon takes place in a river and is an out-and-back course. The river has a steady current that is 0.400m/s and the athletes swim against the current for 750m before turning and swimming with the current for the second 750m. If one competitor can swim at a steady speed of 1.10m/s in still water, how long will their swim take in this river? Compare this to how long it would take in the absence of any current.
My question: I know that for this problem I have to add the relative velocities and subtract the relative velocities when the swimmer is going in the opposite direction, but I am not exactly sure where to start or if there is a specific equation I should use?
My question: I know that for this problem I have to add the relative velocities and subtract the relative velocities when the swimmer is going in the opposite direction, but I am not exactly sure where to start or if there is a specific equation I should use?
Answers
Answered by
Henry
(Vs-Vw)t1 = 750 m.
(1.10-0.4)t = 750
0.7t1 = 750
t = 1071.4 s. = 17.9 min. = Time to swim
upstream.
(1.10+0.4)t2 = 750 m.
1.5t2 = 750
t2 = 500 s. = 8.33 min. = Time to swim
downstream.
t1+t2 = 17.9 + 8.33 = 26.2 min. = Time to swim upstream and back.
t3 = d/V = (750+750)/1.1 = 1363.6 s. =
22.7 min., No wind.
26.2-22.7 = 3.47 min. less with zero wind speed.
(1.10-0.4)t = 750
0.7t1 = 750
t = 1071.4 s. = 17.9 min. = Time to swim
upstream.
(1.10+0.4)t2 = 750 m.
1.5t2 = 750
t2 = 500 s. = 8.33 min. = Time to swim
downstream.
t1+t2 = 17.9 + 8.33 = 26.2 min. = Time to swim upstream and back.
t3 = d/V = (750+750)/1.1 = 1363.6 s. =
22.7 min., No wind.
26.2-22.7 = 3.47 min. less with zero wind speed.
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