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35. M A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velo...

Question

35. M A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe rela- tive t0 the river:

35. M A canoe has a velocity of 0.40 m/s southeast relative to the earth. The canoe is on a river that is flowing 0.50 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe rela- tive t0 the river:



Answers

$\bullet$ A canoe has a velocity of 0.40 $\mathrm{m} / \mathrm{s}$ southeast relative to the
earth. The canoe is on a river that is flowing 0.50 $\mathrm{m} / \mathrm{s}$ east rela-
tive to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

The canoe has the velocity of 0.4 m per second southeast relative to the earth. The canoe is on a river that is flowing 0.5 m per second east relative to the earth. Find the velocity of the canoe relative to the water. Okay, so the velocity of the canoe relative to the earth is equal to the velocity of the canoe relative to the river plus the velocity of the river relative to the earth. So we see V. C. E. Is equal to v C r V r E. So the velocity of the canoe relative to the river is the velocity of the canoe relative to the earth minus the velocity of the river relative to the earth. So the canoe relatives of the earth. It is south east, so southeast is 45 degrees, so we have the velocity of the canoe relative to the earth, is the velocity in the X direction is zero point for co sign of 45 degrees, Which is 0.283 m/s. And the velocity of the canoe relative to the Earth. In the Y direction Is zero point for a sign of 45° to 0.283 years per second. So the river is only flowing east. So if we're looking at the velocity of the canoe relative to the river in the X direction We have 0.283 -0.5, Which is negative 0.217 meters per second. And in the y direction it's just the velocity of the canoe. So then the total velocity is the square root of V. X squared plus V. Y squared. So squaring these two and adding them together, taking the square root, you get the magnitude of the velocity 0.357 meters per second. The angle is the inverse tangent, A. B. Y over Vieques. And we get 52.52° south of West

In this video, we will find the actual velocity of a boat that is traveling at 15 mph relative to a river that is flowing at 3.9 mph with the boat Traveling in the direction of 25° relative to due north relative to the water, and the river flowing in the direction of 135°. The first thing we can do to solve this problem is to draw these two victors, as is done on the graph shown here and this is 25° from due north And 1:35°. And next thing we can do is find the angles here and here. Alpha and beta, where alpha is 90 degrees minus 25 degrees, which is 65 degrees and data is 1 35 minus 90 Which is 45°. Next we can find the X and Y components of both factors using the angles Alvin beta, because we know that the X component is equal to the magnitude of the vector at times cosine of the angle and the Y component is equal to the magnitude of the vector times sine of the angle. So the X component of the top vector, which we can call you okay Is equal to 15 times co signed 65, which is 6.4 And the y component is equal to 13.6. And the X component of the bottom factor, which we will call V is equal to 3.9 times coastline 45, Which is 2.8. And the y component is Also 2.8 and we put that negative there because it is opposite the direction of the Y component of the vector. U. So we can add these two to get the X and Y components of the resultant vector, which gets us 9.2 for the x component and 10.8 from an expert for the y component, Which gives us the vector 9.2 to 10.8 to find the magnitude of this vector. We can use the pythagorean theorem where the magnitude of U plus v is equal to the square root of 9.2 squared plus 10.8 squared, or the square root of the extra credit squared plus the y component squared, which gives us 14 .2 mph. And the formula inverse tangent of the Y component over the X component gives us the angle. And so that is the inverse tension of 10.8 over 9.2, Which is 49.6. However, that is the angle if this is the resultant vector here, for example, that is this angle here and we need this angle because we need a relative to due north. so the angle we need is 90 minus 49.6 or 40.4. Yeah. And so these are final answers, and they describe the actual velocity of the boat.

This problem covers the concept of the galilean transformation and from the galilean transformation, the velocity of the plane with respect to uh the river equals the velocity of the plane with respect to the Vinge plus the velocity of the wind with respect to uh level. Or you can write the velocity of the wind with respect to uh ground plus the velocity of the ground with respect to the river. Or we can buy it along the X direction. The velocity of the pain with respect to the river along the extra channels, the velocity of the plane with respect to wind along the X direction, that is 30 meadows four second. That is along the east direction. Plus the velocity after wind with respect to ground along the east direction Is 0m four seconds. And the river flow along the north direction. Therefore the velocity of the river with respect to ground or the velocity of the ground with respective rivers. Zero. So the velocity of the plane with respect to river, along the extra action is 30 m for a second. Let's say this is the question. The velocity of the plane with respect to the river along the north direction wiles ah The velocity of the pain with respective wind along with no direction is zero. Plus the velocity of the wind with respect to ground is 20 m/s. Okay, and plus the velocity of the ground with respect to the river is Uh 5m/s along with direction. So the velocity of the plane with respect to rivers 25 m/s. So the resultant velocity of the brain with respect to river equals spiralled off. 40 m25 sq are the velocity of the plane with respect to the river equals, yeah, 39.1 m four seconds. And the angle theater made by the resultant velocity with respect to east is given by dan in brasov, 25 m four seconds upon uh 30 m for a second. Or the angle theater made by the velocity vector is uh 39.8° north of east.


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