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Two runners start one hundred meters apart and run toward each other. Each runs ten meters during the first second. During each second thereafter, each runner runs ninety percent of the distance heran in the previous second. The velocity of each person changes from second to second. However, during any one second, the velocity remains constant. Make a position-time graph for one of the runners. From this graph, determine (a) how much time passes before the runners collide and (b) the speed with which each is running at the moment of collision.

A. 6.6 $\mathrm{s}$B. 5.31 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

University of Michigan - Ann Arbor

Simon Fraser University

Hope College

University of Sheffield

Lectures

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and this problem. Two runners are approaching each other at speeds that are exactly the same, but that each second reduced by, um, to amount. That's 90% of their previous speed. So we can consider one runner because each runner, if they're doing the exact same thing, that he's gonna run 50 meters. So that's analyzed one runner instead of the two in the 1st 2nd So from time 01 the runner's gonna go 10 meters in the 2nd 2nd from 1 to 2. The runners going to go 90% of that in the 3rd 2nd from 2 to 3, the runners going to go 90% of that. So I'll take the previous number and then multiply that by 90% which is 10 times 100.9 squared from 3 to 4. It's going to be 90% of that. So I'll just jump right to the X going it. It's gonna be 10 times 100.9 cute. So you start to see that this is a geometric sequence and what we want is the sum of all of these different distances, and we want that some toe add up to 50 so this becomes a geometric series and the formula for this, the sum of a geometric sequence, which is a geometric series. If you remember this like it still, um, is she probably learned it in algebra somewhere. It's the some of the about the of the series is equal to the starting value. A times one minus the common ratio are raised to the end. Minus means are divided by one minus are. So in our case, A is 10. The common ratio was 0.9 and N is the number of seconds that will pass until we get the, um, value that we're looking forward to. 50 so we'll set. The sequel to 50 is equal to 10 times one minus 10.9 to the end, divided by one minus 10.9. No. One minus 10.9 is just 0.1. So I can simplify if d equals tend one minus 10.9 to the end over 0.1 and 10 over 100.1 is 100 so just simplifying further. So then I can divide both sides by 100 and keep scrolling down point, huh? 50 over 100 is 1000.5 equals one minus right nine to the end 0.5 minus 0.1 is negative 0.5 and then I can cancel the negatives. So here I have an expression where I need to sell for the exponents. The best way to do this I'm just going to continue appear because I have all this space. The best way to do this is take the lock of both sides. But I take the log of 0.5 and then I said it equal to the log of 0.9 to the end. I could use the rules of law algorithms to simplify lava 0.5. I'm gonna leave as it is. But for this side on the right, we can take the end and put it in front because that's the magic of clogs. So now I want to sell for end. So it's just gonna be dividing both sides by point night log of 0.0.9. So I have and is equal to the log of 0.5, divided by the law of 0.9, which I just plugged into the calculator and I get a value are 6.6 seconds. So we know that the runners meat or they collide after 6.6 seconds the second part of the problem asks us to find how fast they're going at that point where they collide. So if you look at the because each distance happens in one second, numerically the distance travelled is the same as this the velocity at that point because you're just dividing by the distance by one. So I'm gonna take our formula for the distance and figure out what we get for the distance for each is gonna be 10 times 0.9, race to the end. And if we divide that distance by a second, that will be our speed. So as I plug my number and this is going to be the sixth, it's between six and seven. So we know that between zero and one it would went 10 from 1 to 2. It went less than that. So from 6 to 7 were in the 6th 2nd So my aunt is equal to six. So I do 10.9, raised to the sixth over one second and plugging that into your calculator, you will get 5.3 meters per second and it's gonna be that speed for the whole second until e collide

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