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June 12th, 2019, 02:26 PM   #1
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Unhappy Relative motion in two dimension


In a 300 m wide river, flowing with a velocity of 1 m/s, two boats, one with a velocity of v1 = 20 m/s relative
to the Earth and the other with a velocity of v2 = 15
m/s relative to the river, start moving from point K, as shown in the figure. What will the distance between the boats be when they reach the other side of the river?
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June 12th, 2019, 02:45 PM   #2
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boat #1 ...

$v_{x_1} = 20\cos(37) + 1 \approx 17 \text{ m/s}$

$v_{y_1} = 20\sin(37) \approx 12 \text{ m/s}$


boat #2

$v_{x_2} = 15\cos(53) - 1 \approx 8 \text{ m/s}$

$v_{y_2} = 15\sin(53) \approx 12 \text{ m/s}$

Use the velocity in the y-direction to calculate the time required to displace 300 m for each boat, then determine their respective displacements in the x-direction from position K to determine $\Delta x_1 - \Delta x_2$
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June 12th, 2019, 11:16 PM   #3
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Quote:
Originally Posted by skeeter View Post
boat #1 ...

$v_{x_1} = 20\cos(37) + 1 \approx 17 \text{ m/s}$

$v_{y_1} = 20\sin(37) \approx 12 \text{ m/s}$


boat #2

$v_{x_2} = 15\cos(53) - 1 \approx 8 \text{ m/s}$

$v_{y_2} = 15\sin(53) \approx 12 \text{ m/s}$

Use the velocity in the y-direction to calculate the time required to displace 300 m for each boat, then determine their respective displacements in the x-direction from position K to determine $\Delta x_1 - \Delta x_2$
Answer would be 625 but in my book it is 600
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June 13th, 2019, 05:20 AM   #4
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my mistake ...

Boat #1’s velocity is w/respect to the earth ... that means its velocity in the x-direction already takes the current into account. $v_{x_1} = 16 \, m/s$

So, relative motion in the x direction is 16 - (-8 ) = 24 m/s

25 seconds to cross the river results in 600 m separation.
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