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At 2:00 PM a car's speedometer reads $ 30 mi/h $. At 2:10 PM it reads $ 50 mi/h $. Show that at some time between 2:00 and 2:10 the acceleration is exactly $ 120 mi/h^2 $.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 2

The Mean Value Theorem

Derivatives

Differentiation

Volume

Harvey Mudd College

University of Nottingham

Boston College

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So we're being asked at two PM a cars for governors which thirty miles per hour at two ten PM and reach fifty miles per hour. And then we're being asked to show that at some time between two o'clock and to ten dicks elevations exactly one hundred twenty miles per hour squared. Uh, okay, So what we have to do is we're told this time interval, we're told from two o'Clock to to Canada and that automatically gives us a time. Interval gives us from two. PM two to ten p. M. And that is the interval we are in. And since this car is continuously traveling, it is a philosophy is predictable. And we know that, you know, we don't sponging. We don't just discreetly change speed. We know that the speediest continuous and since the spirit, if we have a speed, we can also calculate its derivative. And I'm sorry about that. And we know that the it's the speed is continuous. It's it's ah, it has an acceleration which is defensible. So we know that is continuous and eventual, which means we can apply. I mean values. So yes, so we can see part of united zero mean very serum here so we can trace every prime us T is equal to be at two. Ten minus the ATU over Q ten minus two. And then, if you recall the prime of tea just the same thing as the acceleration. So the derivative with respect to time of the velocity acceleration. So the ex elevation and then we know that three of Q ten is taught to us. It is fifty miles per hour cannery fifty miles per hour and then thirty miles per hour. And I recall this is two. Ten PM to minus two and it's a ten minutes and central units are different. We have tio, we can actually Ah, we write ten minute. So we know that ten minutes is one sixth of an hour, so we can actually substitute that in. We're going to bring that back. So we're going to say, Hey, that way a of T and equal to twenty miles per hour over one sixth of an hour. And this is a gift. This gives us one hundred and twenty miles per hour square, and you can see that with the units because H is at the bottom you bring this up and this goes to the denominator squares to bottom. And this proved that thick solutions exactly one hundred twenty miles. Problem at some point between two o'clock and two ten.

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