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So, we literally just did change in v, which is that one, delta v over change in t over delta t to get the slope of this line, which was our best approximation for the derivative when t is equal to 16. They give us v of 20. Voiceover] Johanna jogs along a straight path. Let's graph these points here. And then, that would be 30. So, 24 is gonna be roughly over here. And so, what points do they give us?

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AP CALCULUS AB/CALCULUS BC 2015 SCORING GUIDELINES Question 3 t (minutes) v(t)(meters per minute)0122024400200240220150Johanna jogs along a straight path. So, let's say this is y is equal to v of t. And we see that v of t goes as low as -220. We see that right over there. For 0 t 40, Johanna's velocity is given by. So, she switched directions.

Johanna Jogs Along A Straight Path

It goes as high as 240. Fill & Sign Online, Print, Email, Fax, or Download. Well, just remind ourselves, this is the rate of change of v with respect to time when time is equal to 16. For good measure, it's good to put the units there.

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For zero is less than or equal to t is less than or equal to 40, Johanna's velocity is given by a differentiable function v. Selected values of v of t, where t is measured in minutes and v of t is measured in meters per minute, are given in the table above. And we see on the t axis, our highest value is 40. And we see here, they don't even give us v of 16, so how do we think about v prime of 16. So, that's that point. So, when our time is 20, our velocity is 240, which is gonna be right over there. So, we can estimate it, and that's the key word here, estimate.

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We could say, alright, well, we can approximate with the function might do by roughly drawing a line here. And so, then this would be 200 and 100. But what we wanted to do is we wanted to find in this problem, we want to say, okay, when t is equal to 16, when t is equal to 16, what is the rate of change? And so, let's just make, let's make this, let's make that 200 and, let's make that 300. So, if we were, if we tried to graph it, so I'll just do a very rough graph here. This is how fast the velocity is changing with respect to time.

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We see right there is 200. So, the units are gonna be meters per minute per minute. But this is going to be zero. So, they give us, I'll do these in orange. When our time is 20, our velocity is going to be 240.

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But what we could do is, and this is essentially what we did in this problem. And so, this is going to be 40 over eight, which is equal to five. AP®︎/College Calculus AB. So, let me give, so I want to draw the horizontal axis some place around here. If we put 40 here, and then if we put 20 in-between. We can estimate v prime of 16 by thinking about what is our change in velocity over our change in time around 16. And when we look at it over here, they don't give us v of 16, but they give us v of 12. Estimating acceleration. So, we could write this as meters per minute squared, per minute, meters per minute squared. So, that is right over there. It would look something like that. Let me give myself some space to do it. And we would be done. Let me do a little bit to the right.

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So, when the time is 12, which is right over there, our velocity is going to be 200. That's going to be our best job based on the data that they have given us of estimating the value of v prime of 16. So, our change in velocity, that's going to be v of 20, minus v of 12. So, let's figure out our rate of change between 12, t equals 12, and t equals 20. And we don't know much about, we don't know what v of 16 is.

So, this is our rate. So, v prime of 16 is going to be approximately the slope is going to be approximately the slope of this line.