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A Coast Guard ship is traveling at a constant velocity of 4.20 m/s, due east, relative to the water. On his radar screen the navigator detects an object that is moving at a constant velocity. The object is located at a distance of 2310 m with respect to the ship, in a direction 32.0 south of east. Six minutes later, he notes that the object’s position relative to the ship has changed to 1120 m, 57.0 south of west. What are the magnitude and direction of the velocity of the object relative to the water? Express the direction as an angle with respect to due west.

$v_{y}=3.04 m / s$ due $165^{\circ}$ south of west

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Simon Fraser University

University of Sheffield

University of Winnipeg

McMaster University

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

10:12

A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.

0:00

A Coast Guard ship is trav…

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A Coast Guard ship is tra…

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Speed and Direction of a S…

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A boat can travel 2.20 m/s…

07:11

A boat can travel 2.20 $\m…

04:08

A ship's captain sets…

So the question states that a coast guard travels at 4.2 meters per second, due east at a constant velocity and on their radar pops up some sort of submarine or something. And initially an angle of 32 degrees south of east. They see it at a distance of 2000 310 meters, and then some time later, six minutes later, an angle of 52 degrees from due west. They see it once again, singles 52 degrees and it's a distance of 1020 meters away. And so our task is to find, uh what, uh, the magnitude is of the velocity of this object relative to the water, which were your calling the origin as well as the angle it makes. So to do this, the first thing we need to do is figure out what this position vector here is. So to find the vector, we're just gonna call it V here, which, uh is the distance that the submarine has traveled relative to the moving boat. We need to subtract thes two vectors through the 2210 meter vector and the 1120 meter victor. So the easiest way to do this would be to first put these in some sort of ah component form. So to separate the components for the 2310 meter vector, it's going to look something like this. So the X component if we're saying East is positive, the X component is going to be 2310 times co. Sign of 32 degrees. And we know that the co sign of 32 degrees is going to be adjacent over high pot news, and the adjacent in this case is the X component of the vector. So we can solve for that. And we'll get this This, uh, turn and same thing goes for the vertical component it's going to be. Since it's in the negative direction, it will be negative. 2000 310 sign of 32 degrees. So that's for the ah 2000. During a 10 meter vector for the other vector, we can do the exact same thing, so the X component is going to be negative. 1120 times co. Sign of 52 degrees and the white component is going to be, uh, also negative. Negative. 1000 120 times, uh, the sign of 52 degrees. So now that we have these two vectors, we can see a subtracting from each other to get the capital victor be. And when we do this, the resulting vector is negative. 2000 568 0.98 comma, 284 0.80 So this is the position vector that we get capital V relative to the boat that's moving. And so we confined the velocity vector, um, of the object that's moving relative to the boat by dividing this whole vector by 360 which is ah, six minutes, uh, in terms of seconds. And we know that distance divided by time is going to be. Velocity said this makes sense. So when we do this, we get a vector of seven negative 7.13 comma 0.7911 So this is our new velocity vector. And so to look something like this, although it will be extended way farther out than I originally drew so like way for it this way So now that we have this vector, we need to find, um, the velocity vector from the origin that connects this point here and the origin. This will give us the, um, velocity of the object relative to the water. So to do this, we just need to add the vector of the boat that's moving. And this vector is 4.20 comma zero because it's not moving vertically at all. And when we do this, the vector becomes negative. 2.93 comma 0.7911 So this is the vector that is ah, connecting, um, the origin and the, uh, relative speed of the the object. So now that we have this, we can find the magnitude of this specter by taking the square root of the horizontal component of the vector just called D of X squared and added to the vertical component of the vector and square it. And when we do this thing, this goes into here and this goes into here and when we do this, we get our magnitude for the velocity relative to the water is equal to 3.0 three for nine meters per second So now that we have this value, we can figure out what the value is of the angle that it makes with the westward direction. And to do this, we can take the 10 inverse of the are components. So we know this X component here as well as the Y component, assuming that this keeps going up and all we have to do is take the tan in verse of the horizontal component of the vertical component over the horizontal component. So I was going to be 0.7911 over, and we can ignore the negative in this case over a 2.93 tan in verse. And when we do this, we get the angle that's made with the westward direction is 15 degrees north of west, and this is the

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