Calculus Question

[Mark Otto]

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A marathoner ran the 26.2 mile New York City Marathon in 2.2 hours.  Using calculus, show that at least twice the marathoner was running at a rate of exactly 11 MPH (miles per hour).

I have tried this a variety of ways, but I think it has something to do with proving that average velocity equals instantaneous velocity. Either way, it needs to be solved with calculus.

Cany anyone help?

[adrian.wallis]

Maybe this is too simple, but if you find the average speed, 11.91 km/hr, then you know that this will mean that the runner would have to be running at at least that speed or higher for some part of the journey. According to various accelaration laws it is not possible to go straight from one speed to another without going at all the speeds in between. so, assuming the runner started at 0 km/hr you know that he must at some point been running at 11km/hr as he was speeding up at the start. And assuming he slowed down to 0km/hr at the end you know that he must have been running at 11km/hr as he slowed down.

[Mark Otto]

You're right, but it has to be shown using calculus, possibly through the Mean Value Theorem.

[dreamz Veteran]

that's easy. here's the general idea of the theory behind it. you can define the velocity function as a arbitrary function. you know several things:

1. length of time is 2.2 hours.

2. the total distance is 26.2 miles.

you can define the velocity function v = v(t). normalize it so that it starts at 0 and ends at 2.2 hours. draw a curve above the x-axis that crosses at those points. it has to cross the x-axis twice. why? he starts from rest, and when he finishes the race, he slows down and stops.

you know that the total distance is 26.2 miles. since dd/dt = v, you know that the integral of v = distance. thus, the area under your velocity function must be 26.2. we don't have enough information to write a real function, but you get the idea.

so now you have a curve above the x-axis. to reach one velocity, a specific velocity must be passed (intermediate value theorem).

you can fill in the details, but it's easy.

[elliot]

You both said that he must slow down to 0 at the end, but wouldn't he slow down only after he crosses the finish line? I.e. >26.2 :p

Edited January 16, 2005 by elliot

[neowin_hipster]

Assuming that he cannot accelerate infinitely fast and showing that he can't complete it at a speed of 11mph in that time. You can use the intermediate value

theorem. It doesn't get any simpler than that.

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If f is continuous on [a,b] and k is between f(a) and f(b) then there must be a number, c, in [a,b] such that f?=k

[SaiKoR]

You have to assume that the runner starts and finishes at a 0 velocity.

Then in order to cover the distance, the runner must have been running at some speed, s above 11mph since the average speed needed is 11.91mph.

Therefore, since velocity is a continuous function (it cant be anything else) by the intermediate value theorem, there must be at least 2 points on the graph where his speed is 11mph. (once when accelerating and once when decelerating)

You could probably go into more detail, but it has been a long time since I've done anything with the intermediate value theorem...