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