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Find Direction and Angle to arrive directly across initial location?

  • Thread starter Stanc
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  • #1
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Homework Statement


A boat can go 12 km /h in a river that has current of 5 km /h.
At what angle relative to the shore does the boat have to leave to arrive directly across from initial location?


Homework Equations


SOH CAH TOA


The Attempt at a Solution



I just want some explanation of this question. Would my 12km/h be my hypotenuse and the resultant be the straight line across the river perpendicular to the shore or would my 12km/h be the straight line perpendicular to the shore??

Because I get 2 different angles if I use tan-1 (5/12) = 22.16 degrees and sin-1 (5/12) which gives me 24.6 degrees.... Which one do I use?
 

Answers and Replies

  • #2
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No, the 12km/h is the across the river speed and the 5km/h is the river speed so the angle relative to going across the river is tan theta = 5/12. The two speeds add together vectorially giving you a slightly faster speed as you travel across and drift downstream.
 
  • #3
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No, the 12km/h is the across the river speed and the 5km/h is the river speed so the angle relative to going across the river is tan theta = 5/12. The two speeds add together vectorially giving you a slightly faster speed as you travel across and drift downstream.
I do not understand this, If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??
 
  • #4
haruspex
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If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??
Yes. Consider e.g. what would happen if the boat's speed relative to the water were less than the speed of the current. Now there's no way to get straight across.
 
  • #5
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Yes. Consider e.g. what would happen if the boat's speed relative to the water were less than the speed of the current. Now there's no way to get straight across.
So, I should use sin -1 (5/12)???
 
  • #6
haruspex
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Yes.
 
  • #7
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I do not understand this, If I want to end up directly across from my initial point, wouldnt the straight line across the river be my resultant and hypotenuse be my 12km/h??
Sorry, I misunderstood your question and was looking puely at the math of adding the velocities.
 

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