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Problem 49 Hard Difficulty

A runner sprints around a circular track of radius $ 100 m $ at a constant speed of $ 7 m/s. $ The runner's friend is standing at a distance $ 200 m $ from the center of the track. How fast is the distance between the friend changing when the distance between them is $ 200 m? $

Answer

$d \ell / d t=\frac{7 \sqrt{15}}{4} \approx 6.78 \mathrm{m} / \mathrm{s}$

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Video Transcript

so we know that we have a runner and he's sprinting around a circular track at a constant speed of 7 m per second. So we have a person right here running around at 7 m per second. Then another thing we want to keep mind is the fact that this, uh, has a radius of 100 m. So this is also going to be 100 m. Then we have their runners friend standing at a different distance of 200 m from the center. So then we have another 100 right here and the friend, and we want to know how fast is the distance between the friend changing when the distance between them is 200 m. So we want to know that when this is the case right here, when this is 200 m, what is that rate of change? And we have an angle right here and that's gonna become important. So with this in mind, we want to draw this diagram that'll help us somewhat. Um, then we want Thio relate our quantities using the coastline law because we have a lot of signs and one angle. So we see that C squared, which is this C is going to be the distance. So C squared is equal to a squared plus B squared minus two a. B Co Cynthia So using the law of coastlines in differentiating both sides, we get that to see d c d t equals to a B signed theater. Do you say the D team solving for a D. C D. T. Uh, in order to do that, we need to find science, data and data DT eso because it's not given to us, we're going to have to find it ourselves. So since the runner is moving at 7 m per second, we want to relate that to the arc length formula, which tells us that the radius times data is equal to the length. So differentiating both sides again, we'll get that d r d t feta. And this is chain role plus r d theta DT is equal. Teoh de l d T Onley rearrange our values and plug in the ones that we know. We know that d l D t is seven over 100 so we have seven over 100 then we know that that is equal to um, de data over DT. So that's 0.7 radiance per second. So with this in mind, we now need to find data. So we'll go back to our coastline law. I'm in plugging important values, so we'll have. The coastline of Fada is equal to C squared minus B squared, minus a squared over to a B. And when we plug in the values that we know for A, B and C based on our diagram up here, we end up getting at the coastline of Fada is 1/4 um and then we could solve for theta. Or we could use the Pythagorean identities, which tells us that the sine squared of theta plus the co sine squared of theta is equal to one. So just solving this way we get that the sign of data is equal to Route 15/4. So with all these values, now we go back what we had originally, which is when we solve for DC DT and what we end up getting is that this is gonna be a b sine theta d theta DT all over. See playing these values that we've already solved for and that were given We get that D. C D. T is seven Route 15 over four, which means that when the distance between the runner and his friend is 200 m, they're changing their changing at a distance of seven times the route 15/4 meters per second.