Applications

A marathoner ran the 26.2-mi New...

Question

Applications

A marathoner ran the 26.2-mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero.

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Determine the validity of the given statement. A cyclist completed a 60 kilometer race in 3 hours. The cyclist was traveling at exactly 19 kilometers per hour, at least twice during the race, assuming the initial and final speeds are zero. Awesome. So it appears for this particular problem we're asked to take the following statement which once again. A cyclist completed a 60 kilometer race in 3 hours. The cyclist was traveling at exactly 19 kilometers per hour, at least twice during the race, assuming the initial and final speeds are zero, and we were asked to take this statement and prove whether or not it's either a true or B false. So now that we know that we're trying to figure out if the statement is true or false. What we need to do first off is we need to let SFT be the position in kilometers of the cyclist at some time T in hours with the following conditions. So once again, let us know that we're trying to say SFT is our position in kilometers of the cyclist at some time value T in hours, and we're trying to use This piece of information with our following conditions, so we need to note that S is 0, so when T is equal to 0, that means S is 0 is gonna be equal to 0, and when T is equal to 3, so 3 hours. So in S of 3, so S of 3 is gonna be equal to 60, because we know that the cyclist completes their 60 kilometer race in 3 hours. Fantastic. So, therefore, without a doubt, the average speed will be S of 3 minus S of 0, divided by 3 minus 0. So once again, S of 3 minus S of 0, all divided by 3 minus 0, it's going to be equal to 60 divided by 3, which is going to be equal to 20 kilometers per hour, fantastic. And let us recall and note that since the cyclist motion is assumed to be smooth, meaning that SFT is continuous and is differentiable, that means we can recall and apply the mean value the. And this mean value theorem, which is MVT for short, so the mean value theorem will guarantee that there is at least 1 time C in our open interval of 0.3 in parenthesis where S C will be equal to 20 kilometers per hour. So let us state that V of T, which is the velocity, so V of T is equal to S T to denote the speed, and let us note that we are given that the cyclist started and ended at rest, so that will mean that V of 0 is equal to 0 and that V of 3 will be equal to 0. And because V of T is continuous on the closed interval of 0.3, that's what the square brackets mean, a closed interval. So that means that V of T is continuous on this close interval of square brackets 0.3, and it reaches a maximum value of at least 20 kilometers per hour. Which is somewhere within our interval, according to our MVT our mean value theorem, that will mean, as we should recall, the intermediate value theorem tells us that V of T. Assumes every value between 0 kilometers per hour and 20 kilometers per hour. So in particular, let us consider the value of 19 kilometers per hour because that's what we're told in the prom itself. So now we're focusing on our value of 19 kilometers per hour. We're considering that value now. So, on the way from T equals 0 to the time C where V of C is equal to 20 kilometers per hour, and I just note that the speed increases continuously from 0 to 20 kilometers per hour, which will mean That by the intermediate value theorem, there is at least one time T1 parentheses. C, such that we can go ahead and write that V of T1 is equal to 19 kilometers per hour. So that will mean logically that after reaching 20 kilometers per hour, the cyclist must decrease to finish at 0 kilometers per hour at T equals 3 hours. So again, by recalling and applying the intermediate value theorem, there will be at least 1 time, T2, in parentheses C3, where V of T2 is equal to 19 kilometers per hour. So following that logic, therefore, the cyclist was traveling at exactly 19 kilometers per hour, at least twice, or at least 2 different times during the race. Therefore, that means that our given statement without a doubt has to be true. Hooray, we did it. So, once again, our final answer, looking at our problem once again has to be a true. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Applications

A marathoner ran the 26.2-mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly 11 mph, assuming the initial and final speeds are zero.

Key Concepts

Mean Value Theorem

Average Speed Calculation

Continuous and Differentiable Functions

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