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The leader of a bicycle race is traveling with a constant velocity of $+11.10 \mathrm{m} / \mathrm{s}$ and is 10.0 $\mathrm{m}$ ahead of the second-place cyclist. The second-place cyclist has a velocity of $+9.50 \mathrm{m} / \mathrm{s}$ and an acceleration of $+1.20 \mathrm{m} / \mathrm{s}^{2}$ . How much time elapses before he catches the leader?

$t=5.52 \mathrm{s}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

Hope College

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Okay, so in this problem, we're working with two bicyclists that are in a race. Our 1st 1 is the leader and the 2nd 1 is playing catch up. Now we know that the leader is traveling at a rate of 11.1 meters per second and that this is a constant velocity, meaning the acceleration is zero meters per second squared. The racer who's playing catch up is starting at a slower rate of 9.5 meters per second and his acceleration is 1.2 meters per second squared. Now the questions asking us, How long will it take for the person playing catch up to catch the leader? So the time is going to be the same value, but the leader is 10 meters in front of the person playing catch up. This means that however far the leader travels, the person playing catch up has to travel that far, and they have to travel 10 additional meters because they're trying to make up that distance between them. So for both of these equations, we do not know anything about the final velocity. So the Kinnah Matic equation we're going to be using to solve them is that last one there because it does not have the final velocity in it. So let's do the leaders equation first, Um, his ex value is X equals V, not T. So 11.1 multiplied by time, plus 1/2 A T squared. Now the nice thing here is there's no acceleration for the leader. So this is actually the end of that equation, and this gives us a nice term for acts that we can use to substitute in to the person playing catch up. So switching over to the second person's equation, going with the same one the X equals V, not t plus 1/2 A t squared instead of acts. I have acts plus 10 equals the initial velocity multiplied by time plus 1/2 A T squared, which gives us 0.6 t squared. Now, at this point, I'm gonna combine my two separate equations because my first equation one for the leader told me that X is the same thing. Is 11.1 tea so I can take that value and use the substitution principle to put it in there, and that gives us our combined equation. Change colors one more time here of 11.1 T plus 10 equals 9.5 T plus 0.6 t squared. Now, looking at this, we hopefully notice that this is set up more or less like a quadratic equation where we have a T squared value, a t value and one term with no tea at all. So I'm going to rearrange it to set it up to look exactly like a quadratic. We end up with zero equals 0.6 t squared, minus 1.60 minus 10. Now, we're gonna need to use the quadratic formula to solve this. Before I do that, I like to make sure my A term is equal to one. So I'm gonna divide everything by that 0.6 and we get zero equals T squared minus 2.7 70 minus 16.7. And now that it's in kind of that basic quadratic form, we can take this and plug it into the quadratic equation. So that gives us what we follow along with the equation listed here that are X value, which in this case that's well, confusing. Let me change attacks. What we're finding which is our time is equal to the opposite of be so. 2.67 plus or minus radical that b squared minus four A. C. So four times a which is one time see which is negative. 16.7 all over, two times a, which is just too so going through this one step at a time. In our calculator, this reduces to 2.67 plus or minus 8.59 divided by two. And here we end up with two time values, one of which is positive, one of which is negative. We actually only care about the positive value because the time has to happen in the positive part, not the negative part, because he catches him after the timer starts. Not before it starts, and we end up with a time of 5.6 seconds.

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