x, the last place for either of them, 295 meters. And now question A asks us what is the speed or velocity of runner 2 with respect to runner one? Now both of the velocities given to us implicitly are via respect to the ground. And so, and also I wrote that explicitly right here by composing this submanuscript g after the number one, wbelow the subscript g after the number 2. Due to the fact that here our answer to part A will be the velocity of subscript 2 via respect to subscript one. That is the velocity of runner 2 through respect to runner one. And we follow this pattern of adding relative velocities, wbelow if you want the velocity of a through respect to b, which is runner two with respect to one in our instance. a is two and b is one, then you have to erected velocities such that you have actually the initially submanuscript via respect to somepoint, plus the velocity of that exact same somepoint through respect to the second submanuscript you desire. So x in this situation is the ground. And so you wanna have the velocity of runner 2 through respect to the ground which we're offered, plus the velocity of the ground via respect to runner one, which we're not specifically provided. We're given the velocity of one with respect to the ground, but the velocity of the ground with respect to runner one is gonna be the negative of velocity of runner one via respect to the ground. So that’s negative 3.5 meters per second. So this is what we have the right to plug-in for v g one, negative 3.5. And we end up through velocity of runner 2 with respect to one is 4.2 meters per second, minus 3.5. Which is positive 0.70 meters per second. And this positive answer indicates that runner 2 is recording approximately runner one. Since to the ideal is the positive direction in our picture. Okay, and also then part B asks that will end up first? So the final x-place of something moving is it's initial place plus it's speed times time. That’s equation 50 from chapter two. And we deserve to settle for t by subtracting x naught from both sides. And then splitting both sides by v, after switching the sides roughly. And we have final place minus initial place separated by rate. And so the last position will certainly be the complete line, which is at place 295 meters. And for runner one, their initial place is 45 meters. So we have actually 295 minus 45 separated by the velocity of runner one through respect to the ground which is 3.5 meters per second. Which offers the moment of 71.43 seconds. The time for runner two to reach this last position of 295 meters, is going to be only 70.24 seconds. And so runner two will end up initially. Since it takes them less time to gain to this place of the end up line. Now part C asks us, by just how much will certainly runner 2 be ahead of runner one as soon as runner two finishes? And there's a couple ways to answer this; one is you can say, well, how much will runner one go in this amount of time? Because this is the time as soon as runner two finishes. And so wbelow will runner one be at this time? That's not the approach I use though rather I sassist, well, here's the position of runner two via respect to runner one, here's the initial position of runner two via respect to runner one, which is negative 45 meters. Since runner 2 is 45 meters behind runner one. And then add to that the velocity of runner two through respect to one which we uncovered in component A, multiplied by time which is the time it takes for runner two to end up. So that's negative 45 meters plus .7 meters per second times 70.2381 seconds, which is 4.17 meters.">


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This is College Physics Answers through Shaun Dychko. We have a race in which tright here are 2 runners; runner number one is 250 meters from the complete line and also runner 2 is 45 meters behind runner one. And this allows us to create dvery own what their positions are. So the place initially of runner two, we"ll contact zero which would certainly make the position initial of runner one, 45 meters and also the end up line then will certainly be this place x, the last position for either of them, 295 meters. And currently question A asks us what is the speed or velocity of runner 2 via respect to runner one? Now both of the velocities offered to us implicitly are via respect to the ground. And so, and also I created that explicitly right here by creating this submanuscript g after the number one, where the submanuscript g after the number 2. Because right here our answer to component A will certainly be the velocity of submanuscript two with respect to subscript one. That is the velocity of runner 2 through respect to runner one. And we follow this pattern of adding loved one velocities, wright here if you want the velocity of a via respect to b, which is runner two through respect to one in our instance. a is 2 and also b is one, then you need to put up velocities such that you have the first subscript with respect to something, plus the velocity of that exact same somepoint through respect to the second subscript you want. So x in this situation is the ground. And so you wanna have the velocity of runner 2 with respect to the ground which we"re offered, plus the velocity of the ground via respect to runner one, which we"re not precisely provided. We"re given the velocity of one via respect to the ground, but the velocity of the ground through respect to runner one is gonna be the negative of velocity of runner one via respect to the ground. So that’s negative 3.5 meters per second. So this is what we can plug-in for v g one, negative 3.5. And we end up via velocity of runner two with respect to one is 4.2 meters per second, minus 3.5. Which is positive 0.70 meters per second. And this positive answer implies that runner 2 is recording up to runner one. Since to the appropriate is the positive direction in our image. Okay, and then component B asks who will certainly finish first? So the final x-position of something relocating is it"s initial place plus it"s rate times time. That’s equation 50 from chapter two. And we can solve for t by subtracting x naught from both sides. And then splitting both sides by v, after switching the sides roughly. And we have final position minus initial place separated by rate. And so the final position will certainly be the finish line, which is at position 295 meters. And for runner one, their initial position is 45 meters. So we have actually 295 minus 45 split by the velocity of runner one through respect to the ground which is 3.5 meters per second. Which provides the time of 71.43 secs. The time for runner 2 to reach this last position of 295 meters, is going to be just 70.24 secs. And so runner two will complete first. Since it takes them much less time to acquire to this place of the end up line. Now component C asks us, by how a lot will runner 2 be ahead of runner one once runner 2 finishes? And there"s a pair ways to answer this; one is you might say, well, just how much will certainly runner one go in this amount of time? Since this is the moment when runner 2 finishes. And so wbelow will certainly runner one be at this time? That"s not the technique I use though rather I sassist, well, here"s the position of runner 2 through respect to runner one, here"s the initial place of runner 2 via respect to runner one, which is negative 45 meters. Due to the fact that runner two is 45 meters behind runner one. And then include to that the velocity of runner two via respect to one which we discovered in component A, multiplied by time which is the time it takes for runner two to end up. So that"s negative 45 meters plus .7 meters per second times 70.2381 secs, which is 4.17 meters.
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