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anonymous
 4 years ago
Swimming in the ocean a rip current flowing at 2m/s starts pulling you out to sea. You try swimming parallel to the shore at 0.5m/s. What is your actual velocity and direction of travel? If you swim for 30s how much father out to sea (perpendicular to shore) are you?
anonymous
 4 years ago
Swimming in the ocean a rip current flowing at 2m/s starts pulling you out to sea. You try swimming parallel to the shore at 0.5m/s. What is your actual velocity and direction of travel? If you swim for 30s how much father out to sea (perpendicular to shore) are you?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1329141822207:dw From my drawing, u can see that there are 2 forces acting on the man. To find the actual velocity. Calculate the resultant velocity, using Pythagoras theorem. 2^2 + 0.5^2 = F^2 F= 2.06m/s To find direction, tan x = opp/adj = 0.5/ 2 x = 14.4 degrees. After swimming for 30secs , how much further are you from the shore? it is just, 30 secs * 2m/s = 60m Since traveling in the direction away from shore is unaffected by u swimming in a perpendicular direction

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For inverse tan .5/2 I got 1.4 degrees...?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0you can type atan(0.25) in degrees= in the google search window. you get 14.036 degrees. roughly 14 degrees
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