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SSM Two boats are heading away from shore. Boat 1 heads due north at a speed of 3.00 m/s relative to the shore. Relative to boat 1, boat 2 is moving 30.0 north of east at a speed of 1.60 m/s. A passenger onboat 2 walks due east across the deck at a speed of 1.20 m/s relative to boat 2. What is the speed of the passenger relative to the shore?

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$v_{P S}=4.596 \mathrm{m} \cdot \mathrm{s}^{-1}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Cornell University

University of Washington

Simon Fraser University

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.

06:02

Two boats are heading away…

06:31

$\because$ A passenger wal…

05:00

As two boats approach the …

04:08

Two canoeists start paddli…

03:04

03:03

Speed of a Boat Two fishin…

02:16

Suppose a boat moves at 12…

04:11

So the question states that a boat moves at three minutes per second due north, and a person on the boat sees another mode that's moving at a 30 degree angle from, uh, north of east at 1.6 meters for a second. And on that boat, a person is walking due east at 1.2 meters per second. So to solve this problem, let's first draw a diagram of what it looks like. So we're going to call the origin here, um, the water of the ocean. It's not gonna be moving at all. It's not gonna be moving with the boats. Nothing. So it's just gonna be stationary. So the first thing we need to draw is Thiebaud Haute that's moving north at three meters per second. So that is pretty simple. Will just look something like this. It's moving at three meters per second, so now that we have this, we can draw the boat that is seen from the boat that's moving at three meters per second. And so the vector that's gonna come off of that's going to describe this boat scene from the boat. Moving at three meters per second is going to come off the head of this vector here. So it's gonna look something like this. We know this angle here is 30 degrees, so this is going to be 1.6 meters per second as seen from the boat that's moving three meters per second. And on this boat, a person is seen walking due east at 1.2 meters per second. So now that we have all these vectors connected, all we need to find if we want to find how fast the person on the boat is moving relative to the water is the length of this specter here which just this which goes from the origin, which is the, um, stationary water to the person which is moving, which we got from calculating all these relative motions. So to find this, let's call it capital vector V. We can, um, first break up each of these vectors into components. So the first factor, which is just three meters per second in the north direction we can write as zero comma three where this first term is the X coordinate and the second term is the y coordinate were saying East is positive and North's positive. Um, for the second vector that's moving at 1.6 meters per second, we have to break it up into its components. So we know this angle here is 30 degrees and we know it's moving at 1.6 meters per second. So it's pretty simple to figure out what this horizontal velocity is and what this vertical velocity is. We can just use trigonometry. So say the co sign of 30 degrees is adjacent over high partners. So it was gonna be V of X over 1.6. So we know that V of X must be equal to 1.6 times co sign of 30 degrees and the same thing will go for V of why except will you sign instead. So have you going 0.6 sign 30 degrees and these air components that we will put here So 1.6 times co sign of 30 degrees and let me get rid of this seems to be just a little bit bigger and 1.6 times sign of 30 degrees. And then the last vector is just described as 1.2 comma zero. So now that we have all these vectors to get this big vector V. We can add all these vectors together. So at at and when we do this, when we add them all up, we get a vector. One point to push 1.6 co sign Ada Co sign 30 degrees comma three plus 1.6 sign 30 degrees. And this is the vector, which describes Capital V. So now if you want to find the magnitude of this vector, we know that to find the magnitude, it's just thescore air route of the horizontal component squared, plus the vertical component square. And so when we plug in our V of X and R V of Y into this equation, we find that the magnitude of our victor V is 4.596 meters per second, and that's our answer.

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