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The table shows the position of a cyclist_(seconds)(meters)5.5 10.2 17.7 26.1(a) Find the average velocity for each time period: [1, 3] m/s[2, 3] m/s[3, 5] mfsm/sEs...

Question

The table shows the position of a cyclist_(seconds)(meters)5.5 10.2 17.7 26.1(a) Find the average velocity for each time period: [1, 3] m/s[2, 3] m/s[3, 5] mfsm/sEstimate the instantaneous velocity when m/s

The table shows the position of a cyclist_ (seconds) (meters) 5.5 10.2 17.7 26.1 (a) Find the average velocity for each time period: [1, 3] m/s [2, 3] m/s [3, 5] mfs m/s Estimate the instantaneous velocity when m/s



Answers

The table shows the position of a cyclist.

$\begin{array} { | c | c | c | c | c | c | c | } \hline t \text { seconds) } & { 0 } & { 1 } & { 2 } & { 3 } & { 4 } & { 5 } \\ \hline s ( \text { meters) } & { 0 } & { 1.4 } & { 5.1 } & { 10.7 } & { 17.7 } & { 25.8 } \\ \hline \end{array}$

(a) Find the average velocity for each time period: (i) $[ 1,3 ]$ (ii) $[ 2,3 ]$ (iii) $[ 3,5 ]$
(b) Use the graph of $s$ as a function of $t$ to estimate the instantaneous velocity when $t = 3$ .

So there is a common problem where we're given either the position of a rock or a ball or just a random position function and we want to find the rate of change or the velocity or something like that. So let's consider the height function Y equals We'll have 35 minus maybe 11 Cheese squared. OK, so it's nothing specific. Sometimes this is 16 if we're dealing with gravity on earth but really it can be anything and we want to find the average velocity for the different time periods. Well, if we define this as FFT what we can do to find the average philosophy is considered um The time at T equals two. So that's going to be f of 2.1, for example -F of two. And then we would divide that by the difference of the x values And we see that the closer that we get to two. So the closer this gets 20, we get very, very close to our limit, which in this case looks like it would be about -44. But that's just one example. We're going to have many examples where we do the same steps. Just different answers

The position of the cyclist is given on the following table 20. When time is zero seconds, we have serum eaters.

The position of a cyclist is given on the following table. T represents time in seconds and s supposition in meters at zero seconds the position is zero at one second position is 1.4 m at two seconds. Position is 5.1 m, three seconds. We have 10.7 m 40 equals four seconds. The position is 17.7 m and for five seconds. The position of the cyclist is 25.8 m. So human this table of data in part A. We want to find the average velocity for each of the given time period. 13233534 and party. We want to estimate the instantaneous velocity when big when times equal three seconds. Mhm. You're really so we're going to calculate the average velocities here. So we will do a are I That is the average velocity of the cyclist on the interval. 13 So there the average velocity is equal and to fraction Cuban this way. In the numerator, we have the difference between the positions the position at the end time, minus a position at the initial time of the interval Soul, It's going to be s at three, which is the last time on right and point of the interval in that minus the position at the initial time. The interval that is one over. Okay, the difference or the length off. Off time. This case three minus one. So this expression give us you will see the table as of three is 10.7 minus is of one is 1.41 point four in that over to that is nine point 3/2 and that's equal to four point 65. And the units are meters for seconds because the units off velocity is equal to the units off distance over units of time. That is, in this case, meters over seconds. So now in the bar to I, it's interval to three. And there Bye. The average velocity is s at the writing point of the intervals. That is three minus s as the left and point is to. And the time span is three minus two. So we have the disease. It's of three is 10.7 minus is to is 5.1 in that over three ministers one. So we get five point 6/1, that is 5.6 m per second. How far? Three times I is the interval. 35 They're the average velocity given us s at five minus s 3/5, minus three. And that is 25 by eight, minus 10.7 over to which is equal to 15.1 over two, which is equal to 7.55 meters per second. And the final average velocity we want to calculate is on the interval. 34 and that is his s at four. Minus s at 3/4, minus three in that sequel to 17.7, minus 10.7 over one at the seven over one, that is 7 m per second. So we have this four average velocities and we know that the average velocity in terms of geometry is the slope off the line, passing through the points defining the interval. There is the right upon point on left and point. Well, now, in party, we want to approximate or estimate the instantaneous velocity at T equals three in one way. To approximate that is to look at the table and we see that the value. Three, Where we want to estimate instantaneous velocity is the midpoint of interval to four. In that sense, or taking that into account, we can say that a good approximation of the instantaneous velocity at three is the average velocity at the interval to four. So we can say that, right? Given that table three is the midpoint off to four, we can say that the average velocity on 24 is a good approximation. Um, to t instantaneous velocity at T equals three, That is Yeah. Velocity tea with three. It's about average velocity at I'm sorry. On the interval. Yeah. 24 which is when average velocity we having calculated in part A. So this is s at four. Minus s at two. Over four minus two. Uh huh. That's equal to have 17.7 minus is a two is 5.1 over to, and that signal to 6.3 m per second. So the okay, instantaneous velocity off the cyclist when tea was three is about 6.3 m per second. And this in this case happens to be equal to the average off the average velocities on 23 and 34 that is, we calculate the average of thes to average velocities. We obtained just this 6.3 million m per second. So the we have calculated the instantaneous velocity as the approximation given by the yeah, average velocity on an interval. Well, for which this, uh, time this case time equal three is meat point of the interval. And with that, we're having have good approximation off the instantaneous velocity.

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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