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Physics Two boats start from the same dock. The first boatis moving at is moving at 21.0 m/s at 32.00 West ofNorth. The second boat is moving at 35.1 m/s at48.50 N...

Question

Physics Two boats start from the same dock. The first boatis moving at is moving at 21.0 m/s at 32.00 West ofNorth. The second boat is moving at 35.1 m/s at48.50 North of East.A) What is the velocity of the first boat relative to the secondboat?B) How far are the boats apart after 6.0 hours?

Physics Two boats start from the same dock. The first boat is moving at is moving at 21.0 m/s at 32.00 West of North. The second boat is moving at 35.1 m/s at 48.50 North of East. A) What is the velocity of the first boat relative to the second boat? B) How far are the boats apart after 6.0 hours?



Answers

A 200 -m-wide river flows due east at a uniform speed of 2.0 $\mathrm{m} / \mathrm{s}$ . A boat with a speed of 8.0 $\mathrm{m} / \mathrm{s}$ relative to the water leaves
the south bank pointed in a direction $30^{\circ}$ west of north. What are
the (a) magnitude and (b) direction of the boat's velocity relative
to the ground? (c) How long does the boat take to cross the river?

A cyclist, traveling traveling southeast longer road at 15 km/h, feels the wind blowing from the southwest at 25 km/h to a stationary observer, where the speed and directions of the wind, so the velocity of the air with respect to the ground is equal to the velocity of the air respect to the cyclist, plus the velocity of the cyclist with respect to the ground. So finding the angle between the A. C. And V A. G. Since we know V A. G, the A C BCG, so this angle between a c n a g is 31°,, so The direction of the wind would have to be 14° north of east.

In this question. We have this situation, um, toe both living from points p at the same time and travel in the direction shown. So are we need to find the velocity off both a relative to both be and then the time You also need to find the time taken. 4 a.m. b two b uh, 1005 100 ft apart. Okay, so, um, this is a relative velocity problem. Hey, sort off to find the answer. To find a relative velocity. First, we need to write down on the velocity factors the velocity of boats A and B in rectal form. Okay, So based on the given that Graham, uh, we A is 40 kinds. I could sign 60 degrees. I had plus sign 60 degrees jihad. And so this is, uh, 20. I had passed through ST Jihad. Pete second. Okay, then, BB, is it for two. 30 co sign 45 degrees. I had thus sign 45 degrees jihad. And so this is 15 to I had supplies. Jihad sheet second. So the velocity off a relative to be, by definition, is velocity off a minus velocity of B. So putting the numbers together So you have 20 minus 15 to I had class 20 routes three minus 15 to J had. And using your calculator, you get minus 1.21 ahead because 13.4 jihad, Eat four seconds. Okay. So to find a magnitude, you need to find the magnitude. So you take the X component, you square the x component, and you add with square of the white component. Then you take the square it off the some, and you get 13.5 ft per second. And then the direction is attention. The wind component divide by the ex components. You calculate this gets in 4.8 Greece. So this is how the direction looks like. Okay, so this is a velocity off a relative to be, and then this angle is, uh, 84.8 degrees. Okay, then the next thing is to find the time taken. So I'm going to let the time taken for a and B to be 1500 ft putt our wiki. Okay, so, um, we can use the money to a relative velocity. Times time be equal. Set it equal to, uh, 1500. Uh huh. So, um, okay, t would just be 1500. Divide by 13.5. The concrete is you get 111 seconds. All one point. It's five minutes. Okay, so just boxing the answers. So this is the relative, the main two of the relative velocity. Uh, right. The velocity of area to to be. And this is the direction. And then this is the time taken for them to be 1500 ft putt. Okay, And that's all.

For this problem. We have a boat that is traveling on some water and the boat is going 70 72 degrees east of north at 24 miles per hour. And the water is flowing due south. So the boat is headed that direction. The water is pushing it that direction and the actual path that the boat ends up taking his due east. So that would be the true philosophy of boat. Now the first thing they asked us to do in this problem is to find the velocity of the boat relative to the water as a vector in component form. So we just want to take our vector of the boat here and write it using our I N J vectors. And with that, we need to remember that we're doing I and J along the axes. So I was going in the easterly direction. Jay's going in the northerly direction, and what we have is that the vector of the boat he is 24 times co sign can't use the 72 degree angle because the angle we're supposed to use is the angle that we make with the positive X axis with the eye vector, so that's an 18 degree angle call sign of 18 degrees times Vector I plus 24 times a sign of 18 degrees times Victor J. So grab a calculator and we work goes out. Family get 22 0.83 hi plus seven point for two times J. So there's our boat relative to the water now. The second part, they asked just the true speed of the boat. Well, since the boat ends up going in an easterly direction just along the X axis, then that means that vector right there is the true speed of the boat. And that vector is the same as the X component of the boats Original vector. So that means that the actual speed of the boat, the absolute value of the ISS 22.83 miles per hour

They described the path of boat building in some water here for us. And they told us that the boat is going 24 miles per hour and it was going 70 72 degrees east of north. As I've got drawn in the picture here, the water is going directly south, and the first thing they asked us to do is to figure out the vector of the boat relative to the waters. That's why I have the eye in the J Vectors label on here because this boat can be broken down into components using I and J. But the first thing we have to do is realize that we're have this angle down here that we're supposed to use to find the vector of the boat, not the 72 degree angle. So that is an 18 degree angle that we have there. So that means the vector of the boat is gonna be 24 times he co sign of 18 degrees times our vector component I plus 24 times a sign of 18 degrees tons of actor component J. So when we work that out on the calculator, we end up with a value of 22 0.83 high plus seven 0.4 to J. Now. The other thing, they told us wasthe e actual path of this boat is due east, So the boat is going that way. The water is pushing it that way, and the actual velocity, the actual path of the boat is right there. And if that's going due east, then that means that the velocity of the boat is the horizontal component of the boats vector. So that means that the 22.83 now, as per hour, he calls the velocity of the boat the speed of the boat. And if the water, if this punk ends up going due east, that means the water is equal to the vertical component of the boat because those two are making net value of zero vertical motion, the component for the boat would be that way. The component for the water's going down, and if we're going to east, there's no vertical displacement going on in our final vector. So that means that the 7.4 to the vertical component of the vote miles per hour equals the speed of water


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