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Initially traveling eastwvaro (ums north by travelingcircuiam path at uniform speed asthe figure below. The length of LncABC 248and the car completes the turn 42.0W...

Question

Initially traveling eastwvaro (ums north by travelingcircuiam path at uniform speed asthe figure below. The length of LncABC 248and the car completes the turn 42.0What the acceleration when the car located at a angle of 35.00 Express your answer terms of the unit vectors and 0.127 First find the magnitude of the acceleration; then work out the components of the vector: m/s? -0.06 Note that 248 one quarter the circumference of the circular path: mls(b) Determine the car'$ average speed 124/2

initially traveling eastwvaro (ums north by traveling circuiam path at uniform speed as the figure below. The length of Lnc ABC 248 and the car completes the turn 42.0 What the acceleration when the car located at a angle of 35.00 Express your answer terms of the unit vectors and 0.127 First find the magnitude of the acceleration; then work out the components of the vector: m/s? -0.06 Note that 248 one quarter the circumference of the circular path: mls (b) Determine the car'$ average speed 124/21 ms Determine its average ceccleration during the 42.0-5 interval: 0.127 The average acceleration determined from %-vj. m/s? i + ~0.06 You can determine the component of the average acceleration by considering only the components of the initial and final velocities. m/s"



Answers

A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in Figure $\mathrm{P} 6.12 .$ The length of the arc $A B C$ is $285 \mathrm{~m}$, and the car completes the turn in $36.0 \mathrm{~s}$. (a) What is the acceleration when the car is at $B$ located at an angle of $35.0^{\circ}$ ? $\mathrm{Ex}$ press your answer in terms of the unit vectors $\mathbf{i}$ and $\mathbf{j}$ Determine (b) the car's average speed and (c) its average acceleration during the $36.0$ -s interval.

Okay, So this problem is about a car traveling in a circular path and it's shown in the figure I'm gonna stop and look at the figure, okay? And we know that the length of the Ark ABC is 2 35 and it takes 36 seconds to make the turn. So then we have a few questions. First, let's get down the givens, and then we can discuss, actually what the, um, question is so I'll call the Ark linked to this. That's what I remember calling it. Um, when I took some math class, maybe remember the last time I did that, uh, to 35 meters. And the time was 36 seconds. Okay. And let's draw the picture. So, um, ABC and then be it's 35. It's do that a different color. So a is here. B is here, and C is here. And then we know that at B, this angle is 35 degrees in a double truck and great. So our first goal is to get the acceleration when the car is at the angle of 35 degrees and we want it in the unit in the terms of the unit vectors I hot and J a happ. So, um, if the acceleration at this point is going to be a centripetal because it's moving in the path of a circle and it's going to be the velocity squared divided by the radius But we don't know the velocity. What? We can figure it out using this arc length information. So, um so arc length in general, um, is equal to our times data. So, um, this state I would not be this angle, but the full angle that it goes through to go from a to C so that's gonna be equal to, um So I'll write that down. That data is pi over two. We'll call this 1 35 degrees Fi. I'm guessing we'll use it later, and yeah, and then we'll call this Radius R So, um And then, of course, it's, uh the centripetal acceleration is going to be towards the center, so that's where that 35 degrees will come in. But first, let's find the magnitude of it. So us is. Are they, uh, we know if they ah, are we given the radius? So we're not given the radius? I don't think just triple check. Okay. Well, we'll see. So So we have we have our data here. Um, just actually that. And we know that, um, Omega is Let's assume it's constant. It's gonna be the change in data over the change in time. So, um and then we also know Omega is equal to be over our. So now we have a direct connection to this. Be so we can say V is equal to R divided by time times data, and then we know asked is equal to arth data. So you can say asked over time, I guess I would have I was probably simply really think about that. The velocity is just a total distance divided by the total time. But that's okay. You don't have to always be as efficient as possible. Okay, so we know the velocity, and it's, um so we can put that in here. We have s square and divided by t squared times are and you know, we still have this are swim me pause and think how to deal with that. Oh, I see how we can deal with it. Um, looking at this, we actually know ass and then we know State s. So we confined Our our is just equal to s a birth data. Um, so we can go ahead and plug that in right here. So let's do that. So we still have the s squared. We still have the t squared. And then now we plug in Just plain are we're gonna have an ass here and a data here. So we have us data. Cancel one factor of us s data divided by a T Square. And I wonder if there was an easier way to get that. Not Not sure. So now we can start plugging everything in so to 35 times high over two. And we wanted to buy that whole thing by t squared. So 36 squared. And when I put that into a calculator, I got point to a five meters per second squared. Now that's in this direction. So, like, our acceleration vector looks like this. And then if this angle between here and here is 35 um, that's the same thing. Is this angle And so we can say than that the a vector. We could break it into its components. Its point needs 285 and then the horizontal component is in the to the left. By convention, we would say that's negative. And so it's gonna be minus co sign of 35 degrees in the eye hot direction. And then the vertical is gonna be a plus the sign of 35 in the J hat direction and so weaken, distribute that through. So let me go ahead and play got into my calculator. So I'll take that and co sign 35 degree is. And I got 0.233 negative, 0.233 and the I had direction and and then in the other direction we got, we wanna do sign. And then So we got 0.163 in the J hot direction. And then these both have units of meters per second squared. So I'll just put me there for a second square. Did you know that? Okay, And then next we went the average speed solely. Will that be so Average speed is gonna be the distance divided by the time. Oh, that's funny. Maybe there is an easier rate appoint acceleration, cause I already kind of came up with B is us over tea. So that's just to 35. Divided by 36. So let's put that into a calculator. Yes. 2 35 divided row. Not vice six by 36. Yeah, I got 6.53 meters per second. And for a C um, we went the average acceleration during the time interval. So this really highlights that you might say, Well, this feeds not changing. How do we have an average acceleration and acceleration is the difference in the final and initial vector divided by the total time so that we have the total time is just gonna be this tea and 36. But let's think about our XLR velocity vectors. So, um, to our final velocity is ah, the F and we want to subtract our initial velocity. The initial velocity was to the right, So we want to, um we want to add the negative of it. So here is negative v initial. So we want to add these two vectors. Um, so if you put them like, sort of tip to tail, um, and they have the same magnitude, we're assuming it's not changing its speed. Then the vector sum looks something like this, and it's really just gonna be, um this just by, um, trinkets gonna be the square root of two times the magnitude of the velocity. So scored a two times be something we want to divide that by time. So let's put that into a calculator. Um, so my previous answer times the square root of two and then we want to divide that whole thing by, um, 36. So I get 0.256 That's interesting that it comes out different than this cancer. Oh, do we want, I guess we want it in components, Not just the magnitude of it. So we, um So, actually, I guess the components of it are, um you know, here's the X and Y component, So this is its magnitude. So I guess I can change that to magnitude. Um, you fix this statement so that it's true. Um, so that's equal to the magnitude of a and then, um, yeah, And so now, but we want the difference between these. So you basically want take each of these velocities and divided by the time. So the velocity was, um, 25 divided by the by 36. So I get 0.181 um, in the and they're both the same. So yeah, I just took Sorry if it wasn't clear I took this velocity and divided it by time, Um, and then the final. And so, um, I'm like, I think I was clear. I might have just said the whole thing many different times and it might have been confusing someone, like, restart my explanation just to make sure I'm as clear as possible. So here's our final velocity vector. Here is our initial velocity vector and here's are the final sorry, negative initial. And then here's the final minus V initial. We want to take this thing and divided by time to get the acceleration and then we want to do it and component for him. So each component is just gonna be the velocity this velocity divided by time. Um, because he's air equal because the speed is equal. So we'll get that the acceleration is equal to those to the left. By convention, that's negative. So 0.181 again, I just did the velocity divided by time. And that's and that I had direction. And then the final velocity is positive, which my convention or is up, which by convention, this positive. So that's what I get for the acceleration and just go double check my numbers and he sticks and units on there. Yeah, so I

Hi. The car is moving over a circular track. Initially it was moving you east. The direction shoes are here. This is north. And so and he's and a mist. This is the circular track or rich. The car is moving Initially it is moving due east and after on a given time here the car he said to be moving you know the track the circular track is give me was land has been given, as do 35 off meters and it is actually 1/4 off the circum parents off this circular Jack so we can write it like two pi r divided by four. The time given here is 36 10 seconds in which the car goes from point A to point B. So the speech off this car will come out to be distance divided by time. This is 2 35 off meters, divided by 36 off seconds. So it becomes sick Swine by three Neto for a second. No, for first part of the problem, the very first of all find the radius off this circular track which is symbolized by our so to find it used this expression Dubai times off radius divided by four is equal toe to 35 meters. So the radius will come out to be full into two. 45 divided by two times off by So when we put all these values, the radio's comes out We've on 49 0.5 meter. So in the first part of the problem, we have to find the centripetal acceleration that this wind be. Yeah, this point is B and point C is this This is going to see So we have to find acceleration in better form at point B supposed to fall, We will find the magnitude off century Peter acceleration the chance given by the expression we squared divided by our radius. So here it becomes the slide off 6.53 divided by radius which waas 1 49 going fight. So this magnate you off centripetal acceleration comes out to be Zito Wind 285 neater. But the second is square. No, we have to find this acceleration in vector form. So this is the actual direction off this explanation here This angle has been given us 35 degree. So if you take the holes under component this acceleration, it will come out to be, you see cost 35 degree. And if we take its vertical company, it will come out to be a C sign 35 degree. So the horizontal component will be a C X, which is equal to minus S E cost 35 degree because X direction is along negative X axis. So it will be given by minor 0.25 multiplied by cost certified degree, which comes out to be minus zero wine do 33 nita for second square or in vector form, we can say this is minus 0.2 33 I get neater for seconds. Fire. Similarly, a vertical component of this acceleration means component along y axis is positive. You see sign 35 degrees. So we come out to be 0.2 It fi into sign 25 degree or we can see this is plus 0.163 I need a poor second square. So in vector form, this is plus 0.163 Jacob neater. We're sending you square. So overall centripetal acceleration at this point be he's given asked minus 0.233 I care less 0.163 Jacob Neater 70 square. This is an answer for the first part of the problem. No. In the second part of the problem, we have to find the average speed of the car from point A to point C. And this we have found already in the first part. So this averages speech. It's nothing but 6.58 meter for a second. Finally, in part, part of the problem, we have to find the average acceleration. So for this average acceleration, it will be the change in velocity divided by the time changing velocity from point A to point C. So in vector form, it should be we xy minus V and yeah, and divided by the John as the magnitude of the velocity is always 6.5 meters per second. So the velocity at point C will be along. Why access? So we can see this Niecy to be equal toe 6.5 it, Jacob need a per second. And here this is V A, which in vector form and be presented as 6.5 it I get meet up for a second so their difference will be 6.5 it J K minus 6.5 minute. All right, get meter for second. Divided by the time which waas 36 0 2nd So finally this acceleration comes out to be minus zero point 181 Are you here? Blessed 0.181 Jacob, This is the average acceleration from point A to point C in meter per second sweat. This is answer for the third part of the problem. Thank you.

So this question is asking about multiple things. The first thing that this is asking about is to find the speed of the core. So to find the speed of the car, we have X is a good job. Et so X is equal to re T. So we to find me, we have easy, good expertise. So this is X, which is the distance and this is the time and that gives us this. So it won't support me of the problem. We calculate the radius, so that is half to pay your which is 2 35 m And to do half, we have got 15. So the car exhibition a point B is equal to is equal to the square by our res 6.53 square car is 1 50. We do the bad here. Mhm. We get God. So be from part a 6.53. So we have really minus were by T. B. A physical to 6.53 by minus 6.53 Year divided by 36. That gives us minus your queen 181 I Plus your .1 in front. So that is the average acceleration and I hear is that direction. So I that is the direction and J. That is also the direction. Mhm

Solving party of this problem. So here's radius is equal to 50 m and circumference of the circular party is equal to two. Pi R is equal to two pi multiplication. 50 m is equal to 314 metre now 14 top total circumference is equal to 314 by four metre is equal to 78.5 m. Now we all know that V f squared minus V I squared is equal to X So I will use this expression to calculate the value of VF is equal to under hood two x So just putting the value I can write we have is equal to root and the to multiplication to multiplication 78.5 So I get the value of VF approximately equal to 17.7 m per second. This is the answer for party now for part B, The radial acceleration e r is equal to b squared by our so just putting the value I can write 17.7 m per second holy square by 15 m So I get 6.28 m per second square now for party the total acceleration is equal to root under it is quite rural areas square, so just putting the value I can I do it squared plus 6.28 square on simplification. I get the value edge. Just look at it carefully. I get the value. Age 6.59 6.59 m per second, squared at an angle of 17.7 degrees east of south east of south. So this is our final answer.


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