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