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A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar...

Question

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar ($ extbf{Fig. E3.30}$). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) 18.0 m/s to the right? (b) 3.0 m/s to the left? (c) zero?

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar ($\textbf{Fig. E3.30}$). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) 18.0 m/s to the right? (b) 3.0 m/s to the left? (c) zero?



Answers

A railroad flatcar is traveling to the right at a speed of 13.0 m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar ($\textbf{Fig. E3.30}$). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter's velocity relative to the observer on the ground is (a) 18.0 m/s to the right? (b) 3.0 m/s to the left? (c) zero?

So part A here wants us to find the horse a horizontal component of a drops velocity with respect to Earth as well as the horizontal component with respect to the train. Well, since we're told that the, uh that the rain drops are falling directly down the velocity of the drop with respect, Earth will call that VD Sevy in the ex direction or the horizontal direction. So we'll call that V D. E X is equal to zero because the drops are falling straight down, so it's just zero meters per second. So that's part of a rancher for a and now it wants us to find the velocity of the drops relative to the train again in the horizontal direction. So, uh, that's a me actually in the ex direction. So we'll call that VD for drops relative to the train, an ex direction, and that's going to be equal to minus the velocity of the train relative to the earth in the ex direction. Okay, well, that is 12 meters per second, so that in the east direction So this is going to be equal to 12 meters for a second in the West directions all right west inbox it in as the second front of her solution for a question, eh? Now, question B wants us to find the velocity of the raindrops relative to Earth as well as the velocity of the raindrops relative to the train. So the velocity of the drops relative to Earth is going to be equal to the velocity of the train relative to Earth. Divided by the tangent of the angle at which the train is traveling, which is 32 Greeks. So is tangent of 30 degrees, since tangent is opposite over adjacent. So then playing in the values of this expression, we find this is equal to 20 0.8 meters per second weekend box it in is part of our solution will give a new page going here for this last part. So the velocity now of the drop relative to the train is going to be equal to the velocity of the train relative to Earth, divided by the sine of the angle sign of 30 degrees. So we find that this is equal to 24 meters per second and we can box. It is our last part of the solution

So we want to find the magnitude of the essentially the boat relative to the shore, the magnitude of the velocity of the boat relative to the shore and then during after three seconds, what's its position? So we can say we can draw this out this area, right? We can say this. There's a fada to find here. And then there's a angle Fi with the horizontal. This vector right here will represent the boat relative to the shore. This right here will represent the boat relative to the water. Sorry, relatives with water. And then we consider this. Ah, this velocity is the water relative to the shore. We're going to label this as Vector One. This is Vector three and this is Vector to So we can say that the magnitude of Vector three is going to be equal to the square. This the square root of the vector One squared plus a vector two squared all to the one half power. There it is. So that would be these formula. And that would be equal. Teo point one point two squared plus two point two squared. And this is a call to the half power. So the magnitude of the boat relative to the shore is going to be equal Tio two point five one meters per second. So this will be the magnitude. And if you wanted to find the angle, we can say angle is going to be attention to negative one of twenty. Sorry to point to that. Why component divided by the ex component one point two and this is giving us twenty eight point six degrees now. This equals a fada we want to find relative to the shore so we can safe. I simply equals ninety minus data or weaken. Say ninety minus twenty eight point six. And this is giving us sixty one point four degrees again relative to shore. So this will be our final answer for the direction of the boat of the velocity of the boat relative to the shore. And then for part B. They're asking us for the change in displacement and this is going to give us one point two zero comma, two point two point two zero meters per second. And then this will be times Ah, three point zero zero seconds. So we can say that this is going to be equal, though Delta Team Delta D, the displacement is going to be equal to three point six zero comma, six point six zero meters, this would be downstream and then this would be across the way across the river. So at this point, we can say that if you wanted to find the magnitude of this basement, we can say that the magnitude of displacement equals again. Three point six. Where'd plus six point six squared all to the one half power and this is giving us the magnitude of displacement is seven point five two meters and then Seda would be arc. Ten of the are ten of the y component six point six divided by the ex component three point six. And this is giving us sixty. Rather, this is giving us twenty eight point six again. And then, of course, if I would be sixty one point four degrees and if you wanted to get their components, this could be also an answer. That is the end of the solution. Thank you for washing

Our question says that a boat is traveling are a booking. Travel 2.2 meters per second, Still water. Then it says a. If the boat points its proud directly across stream, whose current is 1.3 meters per second, What is the velocity? Both magnitude and direction of the boat relative to the shore and the part VI says. What will be the position of the boat relative to the point of origin after three seconds? So let's first do part a great assets to find the velocity relative, the shore, both magnitude and direction. So I wrote down here what we were given the velocity of the boat. These abuse 2.2 meters per second, the velocity of the water piece of W's 1.3 meters per second and also drew a diagram here that shows the direction of the velocity of the water direction of loss of the boat. Then visa be prime would be the direction of the lost city of the boat relative to the shore. But because of the loss of the boat and the velocity of the water, the dotted line here is the shore. They does the angle of separation between Visa being decent be prime. And FYI, is the angle of separation between Visa be prime in the shore. So we want to find fi as well as the magnitude of visa P. Fine. Okay, so first thing we need Teo, I realise, is that visa be prime magnitude since, uh, we can use both a Grint dirham is going Teo be equal to the square root of the sum of the square on both sides that make it up. So 2.2 Well, it's kind of like this would be more generic. So you're gonna have visa be prime. I'm sorry. Recent b squared. Plus he said w squared. So if you plug in 2.2 meters per second squared for Visa B and B's of W playing 1.3 meters per second, you square it and you add those together. Take the square root, you get two 0.56 meters per second. Okay. And that is the magnitude so we can box it. It was one of the things we're asked to find for part a. Let's indicate that this is part of a okay, then we're also asked to find direction. Well, we can use our triggered a metric identity. The fact that tangent of Fada the angle Fada is equal to the opposite vector, which is VW divided by the adjacent vector, which is Visa Be okay, So solving for theta this gives state is equal to the inverse tangent Greatness tangent to the minus one and we'll go ahead and write the sub w which is 1.30 Looks like sex since fix that There you go. Divided by visa be, which is 2.20 on the meters per second. Cancel out. Okay, so if you take the inverse tangent looks a inverse tangent so tended to the minus one of that ratio, you get 30.6 degrees. But the question asked the angle relative to the shore. So to do that, we need to realize, Well, I'm gonna mark it in red here, so we can seem should this here is 90 degrees. So if we knowthe data that we know fi must be data 90 minus data. Thank you, Degrees. Okay, well, that means that that's 90 minus 30.6, which is 59.4 degrees. So that's the second part of part, eh? So we can go ahead and box that in. Okay? No, for part B again for part B, we're asked after three seconds. So if you give the boat a chance to travel for three seconds full, right, that is T equals 3.0 How far from the origin point would you be now? In order to do that, we are going to first make use of the fact that the distance is going to be equal to the velocity. And these are vectors times the time. Okay. Well, first philosophy this velocity vector B. So you killed a 1.30 All right, plus 2.20 jae ha. And that's in meters per second. Okay, because that's what makes up. Because this V here is Visa. Be prime. Okay, let's be clear on that. And that's visa be prime from the previous. So it's the two sides that make it up. The Serbian be some w. And they're acting in the eye hat. And the jay had directions based upon the X and Y directions. Okay. And then this is multiplied by the time, which is 3.0 seconds. Okay, If you carry out this operation, you'll see this is equal to 3.90 in the eye hat direction plus 6.60 in the J had direction. Okay, Both meters per second. Still, now that we have the two sides that make up the direction, that's not meters per second. This should be in meters because it multiplied with second. So we're in meters now. That's the distance specter. Now you can. Now that we have the distance vector in the X in my direction, we can get the total distance, which is what we want to find. Want to find a total distance away from the origin so we can do that as using the tagging serum where the total distance magnitude is going to be equal to the square root of the sum of those sides. I swear so 3.90 squared plus 6.60 squared right And then that's in meters. To do that, this comes out to be 7.67 meters and relative to the shore. We're still angled away at that 59 degrees, so that's going to be at an angle of 50 nine 0.4 degrees. We go to the previous page. Let's just make sure that's what that angle is that we found for 5 59.4 degrees on DH. That's how far away it's going to be. Mrs. Units meters, Let's abuse their units 7.67 meters away at 59.4 degrees. Relatives ashore.

We are asked to find the average acceleration over a 12th interval for different initial and final velocities. Okay, so for part A, we're told that the, um the initial velocity is to the right or the positive x direction or what? I'm gonna call the positive. I had direction so positive. I hat is positive. Xnegative I had is negative X and we're told that they're moving at 15 meters per second and then they end moving at five meters per second. Well, in general, the equation for the average acceleration Call that a, uh we'll just call it a is equal to, uh, be too the final vector velocity minus V one. The initial vector glossy called Invited by the time interval T Okay, well, here we were told we have V two is five meters per second to the right, so it's positive minus 15 meters to the second. It's also positive since it's to the right and both of these then are in the eye hat direction divided by the time interval which were told his 10 seconds. So we find that this is equal to negative one meter per second squared in the end I had direction. So since it's negative, that means it's moving to the left. That's another way of thinking about that. So you could say negative one meter. Ah, you could also say one meter per second to the left or negative one meters per second. Excuse me, you could say one meter per second, squared to the left or negative one meter per second, squared in the eye hat direction, both of them saying the same thing. Part B. We have the same time interval now, but we're asked to find the acceleration given different initial and final velocities. So in part B, we're told that our final velocity is 15 meters per second to the left, which would be negative 15 meters per second. Right since it's to the left. Now we have to subtract from that. Our initial velocity, which we're told is five meters per second to the left or negative five meters per second. Let's re write that So both of these air in the eye hat direction so we could put I had here all divided by the time interval, which is still 10 seconds. So plugging in these values, we find that this comes out to also equal negative one meters per second squared in the I have direction so you could say one meter per second, squared to the left or negative one meter per second squared in the I had direction and lastly part see, Then we're asked to find the average acceleration but were given different initial and final velocities. So here the final velocity is 15 meters per second to the left or negative 15 meters per second. We subtract from that the initial velocity which we're told us 15 meters per second to the right, so this is just minus 15 meters per second. Both these air in the ex direction or I had direction so that I had on the outside and divided by the time interval, which is still 10 seconds plugging in those values, we find this acceleration is equal to negative three meters per second squared. In the end, I had direction or three meters per second squared to the left box, it in a solution apart, see, and that's the final part of our question


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