Optimizing Boat Direction in a River with Current: A Vector Diagram Approach

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To optimize a boat's direction across a river with a current, a vector diagram is essential. The boat, traveling at 10 m/s, must point slightly upstream to counteract a 5 m/s current. By drawing a right triangle, the hypotenuse represents the boat's speed, while one leg represents the current. Using trigonometry, the sine function indicates that the angle needed to maintain this course is where the opposite side (current) is half the hypotenuse (boat speed). This approach effectively calculates the necessary angle and speed across the river.
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I know i need to use a simple vector diagram ...but I'm not sure how to set it up...

In what direction should a boat point if it has a speed of 10 m/s, and wants to end up directly across a river with a current of 5 m/s.

help? :confused: :bugeye: :rolleyes:
 
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Draw a pictur so that the vector representing your boats speed is pointing slightly up stream (don't worry about the angle now: just label it θ) and has length 10. Draw a horizontal line (across the stream) representing the "course you want to make good" and finally draw a vertical line representing the flow of the stream of length 5.

You should now have a right triangle with angle θ (which you don't know yet), hypotenuse of length 10 and leg opposite θ of length 5.

Do you remember a trig function that is "opposite over hypotenuse"? For what angle is that function equal to 5/10= 1/2?
 
You just have to draw a right angled triangle. The length of the long side represents the boat's speed, and one of the shorter sides is the river speed. The length of the remaining side gives the boat's speed across the river.

By Pythagoras:

5^2 + x^2 = 10^2

x = \sqrt{10^2 - 5^2} \approx 8.66

sin(a) = \frac{5}{10}
 
ahh thanks! I was thinking along these lines but I couldn't decide if the hypotenuse was the 10 or one of the legs..thanks a lot guys :) :biggrin:
 
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