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