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You are a member of a production crew filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite. By watching debris flowing down the river, you estimate that the stream is flowing at 0.653 m / s. In still water, he can swim at 0.719 m / s. At what angle with respect to the shoreline would you advise him to swim to travel directly across the stream to his campfire?

Guest Sep 14, 2014

Best Answer 

 #1
avatar+85894 
+13

Notice the diagram of the situation - I've increased everything by a factor of 10, but the idea is still the same.

 

Let's imagine that, for every 10 seconds he swims, the current is pushing him to the "right" (denoted by segmet CB) by 6.53m while he swims a total distance of 7.19m (denoted by segment AB). And the angle he is being "pushed" away from the "vertical," i.e., his intended direction of "straight across" the river, is given by

sinΘ = (6.53/7.19)   .....thus, to find "Θ" we can use the sine inverse .........

sin-1 (6.53/7.19)  = about 65.26°

So, to counter this, he needs to swm at the same angle on the opposite side of the vertical - i.e. to the "left" side of the vertical. And since you wanted the angle with respect to the shoreline,  then he must swim at an angle of (90 - 65.26)° = 24.74° with respect to the shoreline on his "left.".....Note that the width of the river doesn't matter since, no matter how wide the river is, we still have the same similar triangle !!!!

(I think i'm correct here, but I'm going to have one of our Physics pros look at this one just to see if I've made any false steps!!)

 

CPhill  Sep 14, 2014
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1+0 Answers

 #1
avatar+85894 
+13
Best Answer

Notice the diagram of the situation - I've increased everything by a factor of 10, but the idea is still the same.

 

Let's imagine that, for every 10 seconds he swims, the current is pushing him to the "right" (denoted by segmet CB) by 6.53m while he swims a total distance of 7.19m (denoted by segment AB). And the angle he is being "pushed" away from the "vertical," i.e., his intended direction of "straight across" the river, is given by

sinΘ = (6.53/7.19)   .....thus, to find "Θ" we can use the sine inverse .........

sin-1 (6.53/7.19)  = about 65.26°

So, to counter this, he needs to swm at the same angle on the opposite side of the vertical - i.e. to the "left" side of the vertical. And since you wanted the angle with respect to the shoreline,  then he must swim at an angle of (90 - 65.26)° = 24.74° with respect to the shoreline on his "left.".....Note that the width of the river doesn't matter since, no matter how wide the river is, we still have the same similar triangle !!!!

(I think i'm correct here, but I'm going to have one of our Physics pros look at this one just to see if I've made any false steps!!)

 

CPhill  Sep 14, 2014

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