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And then, when our time is 24, our velocity is -220. It would look something like that. Well, just remind ourselves, this is the rate of change of v with respect to time when time is equal to 16. So, our change in velocity, that's going to be v of 20, minus v of 12. They give us when time is 12, our velocity is 200. For 0 t 40, Johanna's velocity is given by. 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. And so, this is going to be equal to v of 20 is 240. Johanna jogs along a straight path ap calc. We could say, alright, well, we can approximate with the function might do by roughly drawing a line here. AP CALCULUS AB/CALCULUS BC 2015 SCORING GUIDELINES Question 3 t (minutes) v(t)(meters per minute)0122024400200240220150Johanna jogs along a straight path. So, when our time is 20, our velocity is 240, which is gonna be right over there. So, let me give, so I want to draw the horizontal axis some place around here. So, if we were, if we tried to graph it, so I'll just do a very rough graph here.
We see right there is 200. Now, if you want to get a little bit more of a visual understanding of this, and what I'm about to do, you would not actually have to do on the actual exam. If we put 40 here, and then if we put 20 in-between. And so, let's just make, let's make this, let's make that 200 and, let's make that 300. Voiceover] Johanna jogs along a straight path. We see that right over there. When our time is 20, our velocity is going to be 240. And so, these obviously aren't at the same scale. This is how fast the velocity is changing with respect to time. 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? Johanna jogs along a straight path. for 0. 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.
It goes as high as 240. And so, then this would be 200 and 100. And then our change in time is going to be 20 minus 12. So, that is right over there. Let's graph these points here. So, v prime of 16 is going to be approximately the slope is going to be approximately the slope of this line.
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. And we would be done. And we see on the t axis, our highest value is 40. Johanna jogs along a straight path summary. And when we look at it over here, they don't give us v of 16, but they give us v of 12. For good measure, it's good to put the units there. And so, this would be 10. So, this is our rate. So, that's that point. So, when the time is 12, which is right over there, our velocity is going to be 200. So, if you draw a line there, and you say, alright, well, v of 16, or v prime of 16, I should say.
AP®︎/College Calculus AB. Let me give myself some space to do it. Fill & Sign Online, Print, Email, Fax, or Download. And then, that would be 30. And we see here, they don't even give us v of 16, so how do we think about v prime of 16. So, she switched directions. So, we could write this as meters per minute squared, per minute, meters per minute squared. Use the data in the table to estimate the value of not v of 16 but v prime of 16. Let me do a little bit to the right.
They give us v of 20. So, the units are gonna be meters per minute per minute. So, at 40, it's positive 150. Estimating acceleration. So, 24 is gonna be roughly over here. 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. And so, what points do they give us? So, they give us, I'll do these in orange.
We go between zero and 40. And we don't know much about, we don't know what v of 16 is. So, -220 might be right over there. We can estimate v prime of 16 by thinking about what is our change in velocity over our change in time around 16. So, we can estimate it, and that's the key word here, estimate. So, let's figure out our rate of change between 12, t equals 12, and t equals 20. Well, let's just try to graph. But this is going to be zero. And so, these are just sample points from her velocity function.
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.