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The table shows the position of a motorcyclist after accelerating from rest.

(a) Find the average velocity for each time period: (i) $ [2, 4] $ (ii) $ [3, 4] $ (iii) $ [4, 5] $ (iv) $ [4, 6] $(b) Use the graph of $ s $ as a function of $ t $ to estimate the instantaneous velocity when $ t = 3 $.

A.(i) $29.3 \mathrm{ft} / \mathrm{s}$(ii) $32.7 \mathrm{ft} / \mathrm{s}$(iii) $45.6 \mathrm{ft} / \mathrm{s}$(iv) $48.75 \mathrm{ft} / \mathrm{s}$B. $29.7 \mathrm{ft} / \mathrm{s}$

05:46

Daniel J.

01:15

Carson M.

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 1

The Tangent and Velocity Problems

Limits

Derivatives

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So in this problem we are given this table of positions of a motorcyclist after acceleration, accelerating from rest at different times, t in seconds and were asked first to find the average velocity for each time period. 1st 1 being from 2 to 4. Okay, so the average velocity is the average rate of change across the interval. In other words, it's gonna be the Going to be S 2 -11 over T two minus T one right distance over time, which will give us velocity. So From 2 to 4 We have 79.2 -20.6 over four minus two, which is 58.6 over two. So that's 29 0.3 feet per second. Okay, the next interval is from 3 to 4 using the same formula, That's 79.2 -46.5 over four minus three. And so we have 32.7/1. So that's 30 2.7 feet for a second. Okay, the next interval is 4-5, So that's 1 24.8 minus 79.2 over five minutes four. I must four is 1. So this is 40 five. No, this is Yeah, six feet for a second. And then the final period, final interval is from for 26, which we can see from our table Would be 1 76.7 -79.2. So let's see here. 1 70 6.7 -7 79.2 over six months four, Which is 97.5 over two, Which is 48.75 feet for a second. All right, so those are the average velocities. Now. The next thing says to use the graph of S as a function of T. To estimate the incidents velocity when T equals three. So let's go over here and graph this now. So I go to my graphing dez most. Yeah. And enter the data. 012 three 456, wow. And then zero yeah. 4.9 20.6 46.5 79.2 24 0.8 And 1 76.7. So here's our graph of this data. Okay. Here's our graph of the data. So we want to find instantaneous velocity when T equals three. Okay. So what we need to do is is draw the graph through this data. Okay, so let's get the best fit here. That data looks like it is either exponential or a second order parabola happening to us here. Okay, so let's say that why one approximately equal to a X squared plus, but this should be a that's one down there plus B X. Right one plus. See okay, and look at that that data, it's right on that curve, doesn't it? We have a very very good fit our squares .9999. We have a great fit here. Okay, so if I blow this up a little bit. Right, So that I'm here at three. Oh okay. Okay. Well so we know that that the instantaneous velocity at T equals three is the limit as T goes zero over the change in the distance? Over the change in time as well as our horizontal X goes 23. Okay, well we look at our graph here for a second. What can we see about our graph? Oh, we can see that if three is here at 46 a half. Okay then at was at three mm we're looking at this tangent line right of this curve here and so look at this territory are for a minute. Okay then at four. All right, that tangent line would be about where would be about 75, wouldn't it? Okay, so let's use that. So instantaneous velocity Would be 75 46.5 Over four months 3, Which would be 28.5 feet per second. And if you want to be more accurate zoom in closer and use a smaller and smaller interval so that you take this limit right here to get an even more accurate instantaneous velocity

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