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1. Homework Statement
Use the components method to solve this problem.
A river is flowing at 1.75 m/s. The river is 820m wide. You are on a boat that is going dock on the other side of the river, and 940 m upstream. If you need to get to the other dock in 10 minutes, what must the speed of the boat be with respect to the water?
2. Homework Equations
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
3. The Attempt at a Solution
Well i assumed that (as can be seen in my attached picture) the boat was moving in a diagonal direction, so I knew that it would be moving in both the "x" and "y" direction.
x
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
940m= 0+[tex]\frac{1}{2}[/tex](600s)([tex]V_{1}[/tex]+0)
3.13=[tex]V_{1}[/tex]
y
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
820m= 0+[tex]\frac{1}{2}[/tex](600s)([tex]V_{1}[/tex]+1.75m/s)
.98=[tex]V_{1}[/tex]
3.13+.98= 4.11m/s = [tex]V_{1}[/tex]
Can someone please tell me if this method is correct?? If not, please explain. Thanks so much!
Use the components method to solve this problem.
A river is flowing at 1.75 m/s. The river is 820m wide. You are on a boat that is going dock on the other side of the river, and 940 m upstream. If you need to get to the other dock in 10 minutes, what must the speed of the boat be with respect to the water?
2. Homework Equations
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
3. The Attempt at a Solution
Well i assumed that (as can be seen in my attached picture) the boat was moving in a diagonal direction, so I knew that it would be moving in both the "x" and "y" direction.
x
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
940m= 0+[tex]\frac{1}{2}[/tex](600s)([tex]V_{1}[/tex]+0)
3.13=[tex]V_{1}[/tex]
y
[tex]d_{1}[/tex]=[tex]d_{0}[/tex]+[tex]\frac{1}{2}[/tex]t([tex]V_{1}[/tex]+[tex]V_{0}[/tex])
820m= 0+[tex]\frac{1}{2}[/tex](600s)([tex]V_{1}[/tex]+1.75m/s)
.98=[tex]V_{1}[/tex]
3.13+.98= 4.11m/s = [tex]V_{1}[/tex]
Can someone please tell me if this method is correct?? If not, please explain. Thanks so much!
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