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Bicycle Caee sprintsTOCclinch viclory.Faceinilia KelucilyIVs and Jcce ralesLhe rle350 nsWhat hls fInal velocity (In Ms)? 15.45 m{sracer continues a thislocity t the...

Question

Bicycle Caee sprintsTOCclinch viclory.Faceinilia KelucilyIVs and Jcce ralesLhe rle350 nsWhat hls fInal velocity (In Ms)? 15.45 m{sracer continues a thislocity t the tinish line_was 300From the tinish line when he starteccccelerate how Mucre Lime (in s) did he save?otherracerwaz 5,0Dahead when the vinner started seconds did the winner finish?accelerate, but he was unableacceleratetraveled at 13.2 m/s until the finish line. Hov' far ahead him (in meters and indistanceTrc

bicycle Caee sprints TOC clinch viclory. Face inilia Kelucily IVs and Jcce rales Lhe rle 350 ns What hls fInal velocity (In Ms)? 15.45 m{s racer continues a this locity t the tinish line_ was 300 From the tinish line when he startec cccelerate how Mucre Lime (in s) did he save? otherracerwaz 5,0D ahead when the vinner started seconds did the winner finish? accelerate, but he was unable accelerate traveled at 13.2 m/s until the finish line. Hov' far ahead him (in meters and in distance Trc



Answers

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/s and accelerates at the rate of 0.500 $\mathrm{m} / \mathrm{s}^{2}$ for 7.00 $\mathrm{s}$ . (a) What is his final velocity? (b) The racer continues at this velocity to the finish line. If he was 300 $\mathrm{m}$ from the finish line when he started to accelerate, how much time did he save? (c) One other racer was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

So for this problem, you have kidney Matics, Um, basically using motion equations to solve for the various information. The problem you've got identified the givens and the unknowns paranormal for this kind of problem. You told a cyclist sprint at the end of the race at the beginning of the sprints or initial velocity is 11.5 meters per second. She accelerated the rate a of 0.50 m per second squared, and she does that for a time of seven seconds for part A. You're asked for her final velocity, so VF equals V I plus 80. VF is what we're looking for equals 11.5 plus 0.5 for a Times T. It's seven half of 73.53 point five plus 11.5 is 15 meters per second. So that's the final velocity. During the period of acceleration for part B, The problem asks for um, the cyclist continues at this velocity out to the finish line. There's 300 m from the finish line. How much time did she save? So if she didn't accelerate, her velocity would have remained 11.5. So to go that last part of the race. You do t equals X over v. Assuming it was a constant velocity 300 divided by 11.5. You solve that and you get 26 0.1 seconds. That's how long it would have taken if she hadn't have accelerated. But now I'm gonna figure out well, how much of the race that you have left to go. So I'm gonna use X equals V I t plus one half 80 square. You could use almost any equation with X in it because you know enough information, but I'm gonna use this one. That's 11.5 or initial velocity times the time. Seven seconds plus one half a 10.5 times seven squared. You solve this for that displacement and you get 92 0.75 m. So the total displacement once she finishes accelerating is gonna be Delta X. I guess for that part will be 300 minus 92.75 and you get a distance of, um, let's see. Yeah. 300 minus 92.75 is two or 7.25 Yeah, and now you're going to use that? That's how far she's moving a constant velocity, and so we'll continue this up here. T equals X over V again because now it's constant velocity. But now for 207.25 she's going her new velocity 15. And so that time turns out to be 13.8 seconds. So the total time with the acceleration part, I guess for Part two would be the seven seconds plus the 13.8 seconds. So you get 20 0.8 seconds for that second part. So the difference in time would be the time if she hadn't accelerated minus the time given. She did accelerate and you get a delta t of 5.3 seconds. That's answer for Part B. And now for Part C, it says there's another racer who doesn't accelerate. He was or she was in the lead used T equals X over V, that racer with the 300 m at a constant velocity off 11.8 m per second. So it took them 25.4 seconds to finish the race says What's the time difference? So the delta T between the two racers would be the time for them to finish minus the time it took the accelerating bicycles to finish, and you get a difference of 4.6 seconds. Now there's a couple different ways to do the difference of distance between them. That means that the second bicyclist was Accel Er are moving for that amount of time. So the difference between them would be that much time Traveled by the other bicycles is already finished at 25.4 m per second, so X equals V times T 25.4. Okay, times 4.6 you get Oops, Smith. Sorry, Typo there, Back up! We're not moving at that speed yet, but the speed they're moving out. So the second by Sylvester told, was moving at what, 11.8? Yes. So 11.8 m per second times the 4.6 seconds Multiply those two together 11.8 times 4.6 and yet 54.28 54 m. So that should be the difference and distance between them. When the first the winning bicycles crosses the finish

Hello. So for this question, I'm going to be using the graphical method. So you have a But in that accelerates. So this is a velocity time graph. All right. Somebody accelerates with our 3.8. I mean, it's my second squared yeah. And then it accelerates to a certain velocity free. We don't know. And then it maintains that velocity. The total time for the whole journey is uh 7.88 sentence. Right, brother. So this is t We don't know this t Yeah. This after numerous to this velocity ring and maintain the velocity and then the whole time It's 7.88. I want to find the distance traveled here. This region. We want to find the distance that will be the area. What is But we don't know V we don't know t Right? So, we need to find it. Now we know that the total journey, it's uh 50 m. So, the area under this everything right? Everything. The whole graph. So this diagram was like uh trapeze um you know the area of a trapeze um of the truck museum area is given by half Uh air plus three times each. A plus B. Is the two sides. Right? So this is going to be Uh huh. So, The base, the length of the base is 7.88, right? And then cross the other side. This is the other side. This side. All right. And this side is 7.88 minus T. All right. Not that side with a height V. Oh, the height, the height is V. So, the area represents the total distance traveled. Okay, so let's see. Our area would be half. We have 15 corn 76 one of the sea uh Times V. Okay. And everything. Should we go to the total distance? Which is given in the question of 50 meters? All right, let's go to the first part. The second part, Yes. Remember the acceleration for the first uh for the acceleration That's 3.8. That's my second from zero to t. All right. So, we know that acceleration is remind issue over tea Uh initial velocity zero. So it becomes very over tea. Uh celebration is 13.8. So, rather than we We got 3.18, I'm gonna put this back here. All right. So let's see what that gives me. So, uh this vehicle to half times 15.76 months T Then in place of re I'm gonna put time 3.8 c. Uh in post 50. All right. No, I come from the team. So I'm going to multiply through by uh to to get rid of the half. Um What is going to be 3.8 C. Bracket? 15.76- T. You course? 100. Okay so you can multiply through the bracket. Just gonna give away 59.89 t minus 3.8 C. This minus 100 because zero. So this is a quadratic. Sorry this is squared. This is a quadratic. If you find T From a couple later you got T to be 1.90 seconds. and remember remember from this that v. It was three points 18 and you found t to 1.9. Right? So we becomes 3.8 times 1.9. Uh huh. Don't you see you? 7.22. You need a specific So now we know the value of V. Were not the value of T. If you want to find a distance for this area. That's a triangle right here. Of a triangle. We love triangle. It's off time of the base times height, Okay. With us half Times of peace, which is the time. There is 1.9 second tires, The height, which is a video, just 7.22 middle specific. Uh, so this will give you six point, uh, 86 mirrors. This is the distance covid during the acceleration. Thank you very much.

So we have first the diameter of the circular track, I had 100 m and we have a speed of 6.0 m per second. We could we couldn't say velocity rather simply because it is an circular track and velocity is defined in a certain direction. So here we just simply have a constant speed of 6.0 m per second for the runner running around the circular track. We can say that, then the radius is Equalling half of the diameter, of course 50 m. So we considered then the distance traveled for the entire track would simply be the circumference equal. And of course pi multiplied by the diameter and this would be equal to 314 0.29 meters. The time then for one complete rotation would be 314.29 m divided by six m per second. And this is giving us then 52 point 382 seconds. And this uh then would be the time for a one complete revolutionary rotation around the circular track. We can then say the time to travel from A. To B would be 1/4 of this total time. So 52.382 divided by four which is then giving us 13 0.96 seconds. We could then say for the time from A. To C. Would be Equalling half of the total time. So 52.382 divided by two. This would be giving us then 26.191 seconds. And then the time from C. To D. Would be the same the same as the time from A. To B. So this is again 13.96 seconds. Uh After and to answer the last uh for that last time we could say the time from A. B. C. D. Back to A of course would be equaling simply the total time. So just to reiterate again 52 0.382 seconds. So at this point we can figure out the displacement from A. To B. So we can say that the displacement from A. To B would simply be equaling then the square root of R squared plus r squared. This would of course be equal then too are multiplied by radical to this would be 50 m, multiplied by radical. To giving us 70.7 m displacement from A. To C would be equaling to 100 m or two times the radius. Again 100. The distance then from C. To D would be equaling two again 80 X. Times Radical two equaling 70.7 m. And then the displacement from from a back to point A. Again we have a circular track. So the displacement here would actually be able to zero m because you are ending up at your origin. So we can fine then for part A. We know that from A. To B. The X component of the velocity changes from 0 to 6 m per second and the Y component of the velocity changes from six m per 2nd 20 So we can then say that the average velocity part way would be equal to the displacement A. B divided by the time. Maybe. So here this is equally then 70 0.7 m divided by the 13 0.9 55 will round at the very end. And this is giving us approximately 5.4 m per second. We can we can then say that's the average velocity by definition would be equal to the square root of the average x velocity quantity squared plus the average wind velocity quantity squared. We then know that average velocity than equaling radical to times the average velocity in the X direction. And we can then say that the average velocity in the X direction would be equal to the average velocity in the Y direction. And we finally see that then the average velocity in the X direction would be equal to the average velocity divided by radical to so 5.4 m per second, divided by radical to and this is giving us then 3.81 meters per second. So this would be the average philosophy next direction and equaling the average philosophy in the Y direction as well. Now we can write the formula for the average acceleration. So we can say that the average acceleration in the X direction would be equal then then the velocity in the X direction at B mine minus the X velocity at a. This would be divided by the time from A to B. So this is going to be equal to then six m per second. The uh minus zero of course divided by 13.955 seconds. This is giving us 0.46 m per second squared. So this would be the average acceleration in the X direction. We could then find the same thing except for the acceleration of the wind direction. This would be of course V sub y at b minus V supply at a. Here, it's the exact opposite. We're still dividing by the time from A to B. Except here, we have zero m per second minus six, so negative six m per second divided by again, 13.955 seconds. And this is giving us the negative 0.46 m per second squared. So this would be the average acceleration in the average acceleration in the X. Direction and in the Y direction than four part beef. Again, we can define our velocities at point A at point C. Rather, My apologies. And today we have the X. Component of velocity equaling zero. Why component equaling six m per second? This is equally serve as well and this is equally negative six m per second at sea. So we can we know that from A to B, the X component of the velocity increases from 0 to 6 m. And then from B to C, the X velocity decreases from six m to zero 20 m per second. Of course it never becomes negative. So we can then say that the average acceleration in the X. Component uh from A to B equaling then average acceleration, the X. Component of the average acceleration from B to C. And we know that then this is equaling three points 81 meters per second. So we can then say that the X component of the average velocity from A to C would simply be equaling two. These two values uh added to each other with some of these two values divided by two. Of course, that will simply bring us right back to 3.81 year per second. For the why velocity it's a tiny bit different. And this would be from B. C. This is equaling the negative by component of the average velocity for path A. B. So we can then say that the average the white component of the average velocity from A to C would be equal. And then the some of these except that they are uh they have the same magnitude except that they're different, they're different in direction. Uh We can then easily say that this would be equal to zero m per second. So this would be the average philosophy in the UAE average velocity in the X. And we do the exact same thing for path A. C. Now we're dealing with the acceleration so we can see that the average acceleration in the X direction would be equal to the X component of the velocity at c minus the X component of the velocity at a Of course this is equaling zero. This would be of course divided by the time. However, here the numerator again is equaling zero. So this would be equal to zero m per second squared in the y direction we have then the y component of the velocity at sea minus the y component of the velocity at a Again, this here he may be tempted to say zero. However, here it's actually negative six. Keeping with our conventions that we have defined in the beginning negative six m per second minus six m per second. So here we're going from positive six all the way down to zero and then further more going to negative negative six m per second, all in 26.191 seconds. So here we have the why component of the average acceleration Equalling negative 0.46 m per second squared for part C. We want to then define the path from C to D. So here we can say I can. The average velocity would simply be equal to the displacement from CD. However, here the displacement from CD is the opposite to the displacement from A. To B. So this would be equal to negative A. B. And so we can see that then this would be equal to negative 70.7 meters. This would be again divided by the time CD. So the 70 point negative 17.7 divided by 13.0 95 5 seconds. And this is giving us negative 54 m per second for the average velocity from C. Diddy. Again, this is similar to part A. In the sense that then the average velocity would be equaling to of course, Radical two multiplied by the average acceleration rather the X component of the average velocity. And this is again equaling Radical two multiplied by the Y component of the average velocity X component of the average velocity equaling than the average velocity that we defined before, divided by Radical To this is equally then negative 3.81 meters per second. And of course they're the same. So this would also be equal to the y component of the average velocity moving on to the acceleration. Again, we have then that the average acceleration in the X direction would be equal to the X. Component of velocity at d minus the X component of the velocity at sea. This is divided by the time from C two D. And this is equally negative six m per second, divided by 13.955 seconds, giving us negative 0.46 m per second squared. And in the Y direction we have the average velocity average acceleration in the Y direction equaling the the velocity in the Y direction at d minus the velocity in the y direction at sea, divided by again time CD. And this is equally then zero minus negative six m per second. So in fact positive six m per second divided by 13.9 55 seconds. Giving us then positive 0.46 m per second squared. Now we have the round trip. So and excuse me, I'll have to make a small space for part D. Displacement from A to A. Is of course zero. So A B c D E A. So we can immediately say that the average acceleration in the X direction Equalling the average acceleration in the y direction equaling zero. This is the exact same for the acceleration. Again, this is simply because the displacement is zero. And finally we have then the average velocity for party. The average velocity of the runner between A and B. This would be equaling the like an average velocity between A and B. This would be equaling square root of the X. Component of the average velocity from A to B squared plus the why component of the average velocity from A to B squared. This would be then the average velocity from A. To B. Equalling radical to multiplied by 3.81 meters per second. Giving us of course again 5.4 meters per second. So the average velocity is 5.4 m per second. This is not equal to the average speed because of course the total distance traveled is not equal to the net displacement. Of course again this is a circular track. And then for part F we can say that and this goes back to the very beginning where we have a kind of a constant speed. We can say that for part of the magnitude of the velocity is constant. However the direction is constantly changing so he's going so the runner is running around the circle um and at each point in that path the direction of the velocity is changing. However, he is still running at the same magnitude of the velocity or the same speed of 6.0 m per second. We can say that finally velocity is of course, a vector, and speed is a scalar. That is the end of the solution. Thank you for watching.

Here we have a runner running on a circular track. This is a case of uniform circular motion that there is no change in the radial direction as the runner runs around the track and the runner stays on the same plane of the track, that is his path has no component in the access that points outside the plane of the page here. So we will denote that in a diagram for further solution in this problem as follows, well, denote his position, but will denote him rather as a red dot Him or her, I should say. And, um, we're gonna mark that position with a position vector that will call our go. Gonna call that vector Art of Marcus, position the runners position. Okay. The runner also has a instantaneous velocity that's tangential to the path. We're gonna mark that us Another vector veteran. I'm going toe mark in blue and we're going to call this velocity vector vector V. The magnitude of this instantaneous velocity is the tangential speed which is given in the problem of six meters per second. Another critical piece of information that they give us in this problem is the size of the path itself, that is, it has a diameter of 100 meters per second. I'm sorry 100 meters and that's equal to a radius of 50 meters. Let's actually used proper notation here. Diameters 100 meters. The radius is 50. Cho, that's better. The problem asks us to find the X and Y components of the average velocity and the average acceleration between two points and in each part of the problem is a different set of two points that we will consider. So before we get started, let's go over the relevant formula for average velocity and again, velocity is a vector will continue with that notation, but we're gonna denote that it's an average rather than the instantaneous with the subscript a V. This average is actually the difference of the position. Vector of the endpoint will call that our sub f minus the position vector of the runner at his starting point for our survive for initial. And we also need to know because it's a velocity position over time, the final time to traverse that path when he ends up our sabbeth and this is often denote a more compactly, this expression as the difference in position. The different specter over the difference in time for two points. The average acceleration is very similar quantity. Structurally. However, rather than having positions in the numerator, we have velocities in the numerator. So we have vector Visa Beth Final point again. That's a vector minus the initial velocity. And this is again like the expression above over the final time, minus the initial time and the compact way. And I believe this is how it shows up in the book or how they denied in the book is the difference in vector velocity of the difference Vector of velocity over the change in time for the two points in the past to take care of this time because we're going to use the known information about the tracks geometry that it's ah perfect circle of, um diameter 100 meters. We're going to use that information to compute the time it takes this time for a complete lap is the total distance of that lap, which we know from geometry as pi times the diameter Mr. Conference, divided by the speed at which he traverse is that distance we call s for our specific numbers here we have 314 meters divided by our speed return. Tangential speed of six meters per second, 52 and 1/3 seconds. We'll need this lap time for future. Part of the problem the first case acts asks us to compute the average velocity the specific components of which between two points along the path point A this location on the circle to point B. This location these, um just as we did for the runner up above, we can show the position vectors of these points. We're going to call each them R C B and our Sabei respected these with the arrows or both Vectra quantities. Also for in order to compute the average accelerations, we need to show velocities as well here velocities air tangent to the path, going to note them with blue arrows. And I'm going to call them peace be vector and visa. They vector respectably using the formulae we, um, worked out above or looked up and wrote down above, we can compute the average the las thing ab ridge velocity and average accelerations. Now to compute the actual average velocity and average acceleration vectors for average Palat velocity It is a vector quantity and which we're gonna note as the vector with the sub script at a B for average, compute the difference of the position vectors over the change in time for this case from a to B, we have for our sabih the magnitude of the radius of the circle and it is pointing in the positive Y direction for the initial vector are Sabei. It also has a magnitude the radius, but it is pointed in the negative X direction the in terms of the proportion of the total lap. This distance from A to B is 1/4 lap. So because he's going a constant speed, it's going to take 1/4 of the total time to traverse those two points. I'm sorry. That should be a four for 1/4 here of the total time separating these out into component over tea for the X component. We're subtracting a negative, so it remains positive. So we're writing it rather is positive in this case for the y component. It also has a coefficient of four are over tea. It's in the positive. Why direction now a similar process for computing the average acceleration But instead of having position factors in the numerator this time we have velocity vectors. So being a fractional quantity, here we have the final velocity vector. These be It has a magnitude of the tangential speed s. It's pointed in the positive X direction for the starting velocity. The initial velocity is Visa Bay it to has a magnitude of s and it is pointed in the positive Y direction. The same amount of time is required to diverse between these two points each with these respective velocities. As we said above, that's 1/4 of the total lap time. So we'll put that here for the change of time in the denominator writing. These separate components were left with for s over tea in the X direction. And that's a positive quantity. Positive direction we are left with for us over tea for the y component. But since we had to subtract here and why was positive, we're subtracting a positive. We have to put a negative sign there the y direction and that is our average acceleration. Just a sanity check. Let's confirm that these quantities actually makes sense geometrically for a difference in position. We point narrow from the initial to the final. That's position difference. So this vector should have a positive X component and a positive. Why component? According to this, according geometrically and just for clarification sake, let me to note this as Delta are from a to B, does that agree with the value we worked out? Well, it has a positive X component which is going this direction, which is correct. It has a positive why component, which is going upward, which is correct so geometrically things are aligned. What about the case for the velocity vectors? We can't really add them the same way. So I'm gonna put add them together here in a different space. Positive for V A and we got or an upward for a and we got a, um, rightward going for visa Be Let me see if I can't make these what originate correctly here. That's better to dio a vector difference between these. We go from initial to final, as close as we can get it. Yep. And I'm gonna denote that as dealt, uh, the from A to B Victor. So according Teoh, what we computer dear, this factor should be positive going the X direction it is. But it should be down. We're going in the Y direction Negative going. It's pointing downwards. So that is also in line with, um what we computed. Finally, let's plug in the numbers to see what these values be numerically for the case of average velocity are coefficients are for our over tea. This is four times 50 meters divided by 52.3 uh seconds which leaves us with 3.8 to meters per second for each of these coefficients in the, um for the expression for average velocity for acceleration are coefficients are for s over tea which IHS in our case inserting values We have four times six meters per second Vita by our 52 point three seconds we're left with 0.459 zero point four 59 meters per second per second, which is an acceleration quantity. If you wanted new miracle vectors here, you would insert 3.82 for four are over tea You would insert 0.459 for the four s over tea Let's box thes continuing to our next case here We're being asked to find the average velocity going from point A to point C in the book Point A is where it was last time Point sees on the other side of the track. Let's use My style is here. Point sees on the other side, the track that's better. Let's draw some position vectors, okay? And let's labelled an They are vectors. So they get the arrow Super scripts, the your whole head the still a similar procedure for velocities. Just in the case of last time, the initial velocity is upward and the velocity at point C is downward in the Y direction. It's label him. Okay, Now let's do a computation. Same procedures last time. The average velocity here is the position factor at the end point are subsea, which has a magnitude of are disappointed Positive X that, um in point position vector minus the starting point position vector. So our Sabei is also has a magnitude of our and it is pointed in the negative X direction. Last time we saw when we did a quadrant, that was 1/4 of the total lap time in orderto traverse between these points from a to see, this is 1/2 lap so rather Navid. 1/4 in the denominator like last time. We'll have 1/2 of the total lap time for the change of time. Breaking these into components leaves us with ar minus and negative is to our times two. So, again, we have four are over tea for a coefficient in the X direction we have no why component here? Both of our position vectors have no white component, so it is simply a, um, component with zero magnitude. Do. Now, let's look for the average acceleration instead of position vectors, we now have velocity vectors in Yeah, numerator are the velocity vector at the end. Point C has magnitude s speed and has pointed negative y direction. You're going to subtract our initial velocity vet care at point A. It has a magnitude of s and it is pointed in the positive Y direction. There are no X components to either these vectors, same amount of time has passed between them and each of these points 1/2 the total lap time and writing these a separate components, we have a negative y minus A Y leaves us with, um, negative to s. And we're dividing by 1/2 which is Samos multiplying by two. We have negative for s over T for our white component is said before these velocity vectors have no X component. So that is a zero x notice. These are the same values as we computed above. Um, numerically, if you want a new miracle answer in this case, you would insert those in each of these coefficients. We will box, then onward. Next, we want to compute the average acceleration and average velocity vectors for the case of going from point C to point d. Show these on her diagram. Here, Point C is here. Point de is here. Position vectors like last problem point C goes there from the origin to its location at the intersection of the X axis. The, um, position vector for point d goes, uh, downward to their See if I can move this guy much. That's, uh, label them. These were position vectors using our stops using our to denote position decision factor to end point D to notice our city vector. Let's put in the velocities for this problem. Velocity points directly downward from point C for our starting point, and it points in the negative X direction at Point D are in point. Let's mark them, label them plus de vector. See velocity vector de are in point. Let's do the computation, Justus. We did in two parts before our average velocity. In this case, from C to D. It's going to be the position vector at the final point RCD, which has a magnitude of our pointing in the negative y direction. Our starting point is Point C, which, um, we did notice. Vector has a magnitude of our and it's pointed in the positive direction again. This is the difference between vectors. So we have a negative sign subtracting initial from final we the difference between difference in time between initial to final. This case is a perfect quarter lap time, so we'll write that down is 1/4 T. And if I can't make these smaller, make my life a lot easier and go to the next line. So let's try to do this better, Okay? Writing out a separate components. Let's start with the X. We have for oh are over tea and it is negated. It's in the negative X direction now for the y component. It too is negative and it's coefficient is the same. Just this insanity Check our position. Our difference in position vector oops goes from C to D. It should have a negative x component. It should have a negative white component. And indeed it does. So I'm trusting our solutions So far, let's quickly denote that is Delta Oh, from C to D for the case of our velocity Vector fuel. Imagine we take this. See? Put it down here. It's difference would be that orientation which indicates a, um, negative x component, but a positive why component. So let's see if our answer turns out that as we calculate the average acceleration here back to her black pan acceleration average, you do the final velocity. Just going to be s negative going in the X direction. No, let's do that Right. X negative going in the X direction. There we go. Minus the initial velocity has the magnitude of s and it's negative going in the Y direction. It takes 1/4 lap. Time to go between those two points. Just before writing these out a separate components, we will have a negative for s over tea for the X component a positive four s over tea for the y component are geometrically. Our director had that orientation. Um, so that corresponds with the negative going X component and a positive going wide component. So our sanity check once again lands us to believe this. Correct. So let's box thes now will consider the case for going from point A to point A in a full lap. Well, going from point A as an initial point to point is, a final point means that our position vectors are equivalent taking a difference. Um, between equivalent vectors ends up with the zero vector, the no vector as the result in the numerator of our, um, quantity for average velocity. So let's just plug those in as we have in the previous versions. And if we have a no vector of top, that's truck, regardless of how long it took full in. Not in this case, a full lap. We are left with a zero in the for an X component zero for a why component. The same thing applies in the case for average average acceleration are, um, initial velocity and final velocities are equivalent. So we're again left with a zero vector in the numerator, regardless of the amount of time it took to go that full lap. So our acceleration vector is no. Both components are zero. Let's box these next. The problem asks us to calculate the magnitude of the average velocity. In the case of going from A to B, we have that vector from above the average velocity I need to calculate its magnitude. So just to remind us here, so the magnitude here, we need to square each component and then add them. Take the square root of the result we got, which, after a little sleight of hand here we know is four route to you are over tea. Now recall A from the very beginning of this problem the, um, magnitude of our instantaneous velocity not of the average velocity, but the instantaneous felt velocity they gave us as s, which was equal to six meters per second. This, um, magnitude of the average velocity. Now, um s we can write as the circumference over the lap time. There's a conference. We know he is to use my stylist better. Maybe, Maybe so. Maybe not two pi over our over the total lap time Now, if we compare that to a result that we got here for route to over tea times are we can see that four Route two into pi or not equivalent. So for this next part of the question asking if the magnitude of the average velocity is equal to the tangential speed, When you express each of these symbolic terms, we see that they're not equal. Why is this? Well, you'll notice that when we compute an average velocity, we're including on Lee The path boundaries were ignoring all of the velocities that occur along the path in between those boundaries. But by the fundamental theorem of calculus, the closer together that we put our, um, our sit by and our sub f position vectors along that curved path the closer, the resulting average, the value of the resulting average velocity will be to the actual tangential speed average velocity, magnitude. Um, for the case of 1/4 quarter lap, I computed Teoh be around 5.8. I'm sorry. 5.4 meters per second, which is a little bit less than the given value of tangential speed of six meters per second. But that is at least closer than the case of the going from Point A to Point C, which was the um, half lap case that one was merely the 3.82 meters per second she's getting further away. In our case of a full lap, Um, that was one full lap recall that we had zero meter per second, um, tangential speed. So the shorter the distance between the, um, starting point in the end point along your path the closer at least by this inductive argument here, the closer that magnitude will be to actual final um, part of this problem asks if how is velocity changing if the tangential speed is kept constant? And that's because even though the runner is running at the same speed along this circular track as he goes around the track, his position, his direction, the direction he's facing at the direction, um, that he's running gradually changes and, um, velocity changes not only can be changes in magnitude, they can also be changes in in direction, or they can be both changes in magnitude and direction. And for our case, it was just a directional change as the runner goes around his the circular track so well, quickly write this as because of directional change. No


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