Running pace Explain why if a runner completes a 6.2-mi (10-km...

Question

Running pace Explain why if a runner completes a 6.2-mi (10-km) race in 32 min, then he must have been running at exactly 11 mi/hr at least twice in the race. Assume the runner’s speed at the finish line is zero.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says a cyclist starts from rest and completes a 15 mile race in 45 minutes. Suppose the cyclist's speed at the finish line is 0. Explain why the cyclist must have been riding at exactly 20 MPH, at least twice during the race. We're given 4 possible choices as our answers. For choice A, the cyclist must have crossed the average speed of 20 MPH at least twice since the speed is always equal to the average speed at every moment during the race. For choice B, the cyclist must have crossed the average speed of 20 MPH at least twice since the cyclist accelerates at the start past 20 MPH and decelerates past it again to reach zero speed. For choice C, the cyclist must have crossed the average speed of 20 MPH at least twice because maintaining a steady pace is required except at the start and the finish. And for choice D, the cyclist must have crossed the average speed of 20 MPH at least twice because the speed increases linearly until the midpoint and then decreases linearly to zero. Now, we're asked to explain why the cyclist must have been riding at exactly 20 MPH, at least twice during the race. So, the first thing we're going to do is actually calculate our average speed. So, we call for our average speed, which we'll label as V, but that is going to be equal to our total distance, which we'll label as D, divided by the total time, which we'll label as T. Now we're told in the problem that the total distance of a race was 15 miles, so we'll have 15 miles. Divided by the time it takes to complete that race, which is 45 minutes, and 45 minutes is 3/4 of an hour. So we'll divide the 15 miles by the quantity of 3 divided by 4 hours. And so, evaluating this, this is going to be equal to 20 MPH. So that's going to be our average speed. Now we want to look at our position as a function of time. And so we'll label that quantity as S of T. And we know two values for this function. We know its value when T is equal to 0 at the start of the race. So that means that S of 0 is going to be equal to 0, since I have not changed my position at the beginning of the race. And we also know our position at the end of the race. So that's when T is equal to 3/4 of an hour. So S of 3 divided by 4 is going to be equal to 15, since the race here is 15 miles long. Now, our function S of T here is going to be a continuous function, it's going to be a smooth function, and it's going to be differentiable everywhere over the interval going from 0 to 3/4 of an hour. And that's because in real life, our um position is a smooth continuous curve, and we don't really have any abrupt changes in our speed. That means it's also differentiable. So, we have all the properties that we need to actually apply our mean value theorem. And so we call for the mean value theorem that there is a time in this during this phrase, which will label as T1, such that S of T1. is equal to the quantity of 15 minus 0 in quantity divided by the quantity of 3 divided by 4 minus 0, which, when we evaluate that expression, is equal to 20 MPH, which is our average speed. So, what this is saying, there is at least one time during the race at which our cyclists had an instantaneous speed of 20 MPH. Now, since our cyclist started out at rest, that means it started out at 0 MPH and eventually reached the 20 MPH, but in order to have an average speed of 20 MPH, we'll have to pass that um speed in order for the average to be equal to 20. So we'll have to have a speed that's greater than that at some point during the race. We also are told that we crossed the finish line with a speed of 0. So that means that we had to go from that speed that was above the 20 MPH point, back down to zero, again crossing the 20 mile per hour mark. So we've had to cross that 20 mile per hour mark at least twice due to having to accelerate at the beginning of the race, as well as to decelerate at the very end. So based on that information, we can see that the correct answer to this problem is answer choice B. That the cyclist must have crossed the average speed of 24, 20 MPH at least twice since the cyclist accelerates at the start past 20 MPH and decelerates past it again to reach zero speed. So, I hope that this explanation has been useful, and I'll see you in the next video.

Running pace Explain why if a runner completes a 6.2-mi (10-km) race in 32 min, then he must have been running at exactly 11 mi/hr at least twice in the race. Assume the runner’s speed at the finish line is zero.

Key Concepts

Average Speed

Mean Value Theorem

Instantaneous Speed

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