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PreviousProblem ListNextAttempt ResultsEnteredAnswer PreviewResult7.725667.72566incorectThe answer above is NOT correct:points) A car drives down road in such a way...

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PreviousProblem ListNextAttempt ResultsEnteredAnswer PreviewResult7.725667.72566incorectThe answer above is NOT correct:points) A car drives down road in such a way that its velocity (in mls) at time (seconds) isv(t) = 2t1/2 + 5.Find the car's average velocity (in mls) between t Answer5 ad t = 10Preview My AnswersSubmit AnswersYour score was recorded You have attempted this problem time

Previous Problem List Next Attempt Results Entered Answer Preview Result 7.72566 7.72566 incorect The answer above is NOT correct: points) A car drives down road in such a way that its velocity (in mls) at time (seconds) is v(t) = 2t1/2 + 5. Find the car's average velocity (in mls) between t Answer 5 ad t = 10 Preview My Answers Submit Answers Your score was recorded You have attempted this problem time



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The velocity of a car, in $\mathrm{ft} / \mathrm{sec},$ is $v(t)=10 t$ for $t$ in seconds, $0 \leq t \leq 6$ (a) Use $\Delta t=2$ to give upper and lower estimates for the distance traveled. What is their average? (b) Find the distance traveled using the area under the graph of $v(t) .$ Compare it to your answer for part (a).

So first you want to start off by graphing, Um, this model that we have, it's s equals 4.5 t squared. This is our graphic. And we want to know the average velocity over the interval from 0 to 12. So if we look at T equals zero, we see the S zero equal to 12. We see the SS 6 48. So if we take 6 48 divided by that 12, we see that the average velocity is 54 ft per second. Then what we want to do is find the instantaneous velocity of the car At T equals six. So to do this, we want to take the derivative of our initial function. Um, and this is going to be at T equals six. So, um, we'll take the derivative of the initial function. Um, to do this will multiply by the exponents. So it'll be nine p and then no longer squared 19. And then when X equals six, we see rt goes six. In this case, we have to equal six. We see the S equals 54. So in both cases, the average velocity and the instantaneous velocity at T equals six and 12. Um, in each case, the velocity was 54 ft per second

This time we're gonna be looking at Debra. Q. Is applied calculus fifth a dish addition? Chapter five, Section one. Problem number 11 Uh, this problem's interesting. This is going to step up the game conceptually, especially if you've been doing all the problems up to this point, so it's read through it. The lost of your car is gives us a function. F is a function of tea is five team meters per second. Use a graph of FT to find the exact distance traveled by the car and meters from T equals zero t. Who's 10 seconds and you can see have highlighted the word exact. That's sort of what stepping things up here, the previous exercises evolving estimates right. If we take a velocity curve and we break that curve, then we draw rectangles. On top of that. The rectangles are always approximate of because the actual area of the rectangle doesn't perfectly fit, uh, on that curve or under that curve. So let's see if we can get a sketch of this guy and see what's going on. So it's got some axes going t half of T and 10 seconds here. So if I plug in zero. I get zero. Also. Got a point here. If I plug in 10 I get 50. So I have a point out here, And if you remember, functions like this always make straight lines. So I'm gonna draw the straightest possible, and I can cause that's kind of important. Okay, so remember, we've been interpreting the distance traveled by an object. When we're given the velocity curve, the distance is being interpreted as the area under that curve when we've been approximating that area using rectangles. The difference here is because of the geometry of the situation. Does curve is completely straight for shore, right? Five t We know that that makes a straight line, which I guess is a little redundant, since lines are always straight. And so if I want to, I can come up here and we can just ask, Just ask ourselves, Can we exactly compute the area that's under this curve between zero and 10? And the answer is, yeah, because this is ah, triangle right. Geometry class tells us how to find the area of a triangle. Right? We know the area of a triangle is always 1/2 times the base with of the triangle multiplied by a tight, no matter how it's oriented as long as you have one of the sites. Ah, flat for the base. And the high is 90 degrees to that. Uh, that'll be the area. So plug it all. And so we got 1/2 times the base here is going to be linked, Tan, because we're going from zero to towns graph and the height here is gonna be 50 because this dotted line is completely vertical, right? 92 the access and ah, this point was topped out at 50. So if we compute that half tennis 55 from safety is 200 15. Let's check units were quick, The velocity was in meters per seconds. The time was in seconds. That's good. The seconds are canceling. So this is in meters and this is important because this is not an approximation anymore anymore. If you truly had a car whose velocity was always exactly according to this function in the distance, I would travel over this 12th interval would be exactly 250 meters. This is not an approximation, um, important thing to think about on just to leave you with some food for thought. We were able to do this because we knew the exact formula for the area of a triangle. So maybe this, uh, shed some light on. Why? When we have non, uh, linear curves, right. Maybe I'll draw it up here in a different color. Maybe we have a velocity curve. It does something like this, and we want to find the area underneath that. Why? It's a particularly difficult problem because, you know, how are you going to compute that area? Exactly? There's not a easy known formula for that. OK, something to think about.

So this would be the, uh, plot of the the position versus time graph. And we see the parabolic function here. This is Ah, we have, ah, function of time, race to the third power. And so that's what's giving us this. Ah, great cur. There's a exponential curve here. And so we can say that four part Ah, a continuing on for party. We can say that. Um well, these points are simply found at one second to two seconds and three seconds. And then you can simply connect the points in order to sketch the graph eso again, just with the function equaling except T equaling 5.0 meters per second multiplied by tea. And this would be a plus 0.75 meters per second cube multiplied by t cubed Ah, you can simply plug in 12 and three Ah, for tea and find the these points for one. It should be. This would be 5.75 here. This should be 16. And here this should be 35 0.25 So, after plug in, you should get those values and then you can simply sketch for part B now and wants you to find the instantaneous velocity. This would be at T meets a instantaneous velocity at T equaling 4.0 seconds were first using time intervals of 0.4 seconds. So we can say that. Ah, we can calculate ex at T equaling 3.8 seconds and ex at T equaling 4.2 seconds. Again, we're simply using the, uh, formula here, and I'm gonna lose the units so we can. Ah, so it could be a little bit quicker too. Calculate. So five times 3.8, this would be plus 0.75 times 3.8 Cute and then 44.2 seconds. It would be five times 4.2 plus 0.75 times 4.2. Again cute. And so, finding the instantaneous velocity, it would simply be the difference between the the change of displacement with respect to time. So this would be equaling 76.6 meters minus 60.15 meters, and then this would be divided. My 0.4 seconds. And so the instantaneous velocity is gonna be equaling approximately 41.1 meters per seconds. Again, AT T equals 4.0 seconds. Now we're going to be doing the exact same thing. So this would be your answer for a time. Interval of 0.4 seconds. Now we're gonna do the exact same thing. However, our time intervals now 0.2 seconds. So here we have ex at T equaling 3.9 seconds. I'm an ex at T equaling 4.1 seconds. So this is gonna be equaling again. The exact same thing. Five times 3.9 plus point 75 times 3.9. Cubed again. Uh, brother, five times 4.1 plus 0.75 times 4.1 quantity cubed the instantaneous velocity. Is that gonna be equaling 72.2 meters minus 64 meters. I get it simply All right. T X of tea at four point x of tea equaling 4.1 seconds minus x of tea equaling 3.9 seconds. So is simply the difference. And then here, of course, before 0.1 seconds, minus 3.9 seconds. Of course, this is equaling 0.2 seconds, and so this is equaling 41 meters per second. So this would be your answer for part B for a time interval of again. Ah, 0.2 seconds. So delta t equal in 0.2 seconds. Now finally, point. Uh, we have 0.0.1 2nd And so this is where it gets a bit more accurate. Where? Ex at t equaling 3.95 seconds and then ex at T equaling 4.5 seconds. So this would be equaling again. Five times 3.95 plus 0.75 times 3.95 toothy third power. All right. 4.5 plus 0.75 times 4.5 to the third power. And the instantaneous velocity is equaling here. 70.1 minus 66.0. Divided by here. Ah, 4.5 minus 3.5 And this is equaling again. 41 meters per second. So this would be the instantaneous velocity. At four seconds, we're getting closer and closer. So 41 we use per second is essentially pretty close to the exact answer. And so we can say that for part C to find the average velocity, this would be equaling ex At T equaling 4.0 seconds minus x AT T equaling zero seconds and then this would be divided by 4.0 seconds, minus zero seconds. And so this is equaling five times four plus 0.75 times four cube. This would be divided by four. And this is equaling 17 meters per second. So you can see that this average velocity is much less. 17 meters per second is much less than 41 meters per second. And so the average velocity Ah, throughout the entire time interval of 4.0 seconds, or rather from zero seconds to four seconds. This is gonna be ah lot less than the instantaneous velocity at four seconds. So, again, this is taking into account your total displacement from the beginning to the end and how long it took to get there. This is going to be the tangent line at that point in the curve. So with the parabolic functions such as this one, of course, we're going to get a higher, instantaneous velocity. The greater the more time has passed, then if we calculate the entire average velocity from the beginning, So again, the average velocity from zero seconds to four seconds is much much Wes the instantaneous velocity at four seconds, that is the end of the solution. Thank you for watching

So we're told that the distance of a particle from some fixed point can be given by the equation. S A T is equal to t square plus five t plus two. We're t is measured in seconds, and we want to find the average velocity of the particle over the time intervals going from 4 to 6 and 4 to 5. And then they want us to find the instantaneous velocity off the particle when time is able to four. So one of things will need to remember for this question is that velocity is equal to the change in our position over our change in time. So that's why we can use this average rate of change equation on S a T here. So we just need to plug everything in to our equations here. So for the average rate of change for the velocity from 4 to 6, this is going to be s of six minus s up, four over six, minus four. So it's got unplugged these in. So we're gonna have s of six. And this is equal to so 36 plus 30 plus two and adding all of that up that looks like you to give us 68. So So go ahead and replace this with 68 ends and s of war is going to be 16 plus 20 plus two, which looks like it gives us 38 so we can replace that with 38 there. So now if we were to subtract everything, So 68 minus 38 is 30 and then divided by two. That would be 15. And then what were the units were this to see if they tell us? Oh, so they don't actually say the units. We really can't put anything with it. It looks like So I was gonna say 15 and then over here from 4 to 5. So we're gonna follow the same thing that we did before. So it's gonna be s of five minus s for over five minus four. So we already know that s of four is 38. So we just need to figure out what is us of five. So us of five. So it's gonna be 25 plus 25 plus two. So that's 52. So then we would do 52 minus 38 which is 14 divided by one. So just be 14. That would be the average rate of change, or B. Now they want to find the instantaneous rate of change. At times you go four seconds. Well, we could just go ahead and follow this equation. We have appear, and I like to do this in steps. So the first thing you need identifies that are a is going to before because that's where we're interested in with the institutions rate of change. So I like toe first, go ahead and do s of so before, plus h. So it's going to plug four plus h into here So before plus H squared plus five times for plus H plus two. So expend this out. We should have 16 plus eight h most H squared plus 20 lost five h plus two. Now we can go ahead and at all the subs, we'd have 16 plus 20 plus to which is going to be 38 and then we're gonna have eight h plus five h that's gonna be plus 13 h and then lastly, we just have the H squared term right now. What we want to do for our next step is due s for plus H minus best of four. So it's for about what s a four is actually already found that over here we found s a four supposed to be 38 so we could just go ahead and plug that in directly, so there's gonna be 38 plus 13 h plus H squared minus 38 knows thes 30. It's just counts out with each other and were left with 13 h plus h squared. And then for the the last step, we're going to take the limit as h approach zero of s four plus H minus s of four all over h and kind of as a pro tip. Once you get to this point here, if, for whatever reason, whatever you plug into the numeric cannot be evenly divided by this agent the Dominator, then you may need to go back in check your ouch broke. Because at least when we're working with polynomial, they should always cancel out, and so knows here we're just left with 13 plus h, which we can apply this summer directly. Now that'll be 13 plus zero, which gives us 13. So we end up with that instantaneous, falsely being 13


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