Johanna Jogs Along A Straight Path Summary

Tuesday, 30 July 2024

They give us v of 20. And so, this is going to be 40 over eight, which is equal to five. It goes as high as 240. 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

And we would be done. So, -220 might be right over there. And so, these obviously aren't at the same scale. We go between zero and 40. And we don't know much about, we don't know what v of 16 is. For good measure, it's good to put the units there. Johanna jogs along a straight path. for. And so, these are just sample points from her velocity function. It would look something like that. 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. So, this is our rate. 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. For 0 t 40, Johanna's velocity is given by.

So, we can estimate it, and that's the key word here, estimate. Estimating acceleration. And then, when our time is 24, our velocity is -220. But what we could do is, and this is essentially what we did in this problem. And we see here, they don't even give us v of 16, so how do we think about v prime of 16. And so, this is going to be equal to v of 20 is 240. Johanna jogs along a straight patch 1. 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.

Johanna Jogs Along A Straight Path Wow

They give us when time is 12, our velocity is 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, 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 we see on the t axis, our highest value is 40.

We can estimate v prime of 16 by thinking about what is our change in velocity over our change in time around 16. 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. 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. So, let's figure out our rate of change between 12, t equals 12, and t equals 20. We could say, alright, well, we can approximate with the function might do by roughly drawing a line here. Johanna jogs along a straight path wow. So, that is right over there. Let me give myself some space to do it.

Johanna Jogs Along A Straight Path. For 0

When our time is 20, our velocity is going to be 240. So, at 40, it's positive 150. So, they give us, I'll do these in orange. So, we could write this as meters per minute squared, per minute, meters per minute squared. So, our change in velocity, that's going to be v of 20, minus v of 12. And when we look at it over here, they don't give us v of 16, but they give us v of 12. So, the units are gonna be meters per minute per minute. Use the data in the table to estimate the value of not v of 16 but v prime of 16.

This is how fast the velocity is changing with respect to time. AP®︎/College Calculus AB. So, let me give, so I want to draw the horizontal axis some place around here. So, she switched directions. So, when our time is 20, our velocity is 240, which is gonna be right over there.

Johanna Jogs Along A Straight Patch 1

Let me do a little bit to the right. Well, let's just try to graph. And so, this would be 10. So, 24 is gonna be roughly over here. We see that right over there.

And then, finally, when time is 40, her velocity is 150, positive 150. So, when the time is 12, which is right over there, our velocity is going to be 200. So, if we were, if we tried to graph it, so I'll just do a very rough graph here. And so, then this would be 200 and 100. If we put 40 here, and then if we put 20 in-between. So, that's that point. 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. And then, that would be 30.