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