5

Meter / second.Note that if you enter the values expression, 15 + 5 and operations into your calculator in exactly the same orde 3 your calculator wll display the w...

Question

Meter / second.Note that if you enter the values expression, 15 + 5 and operations into your calculator in exactly the same orde 3 your calculator wll display the wrong answer tor the expressior precedence ol operation errorl An alternative approach in case like this is t0 place parenth numerator and denominator separately in your calculator: (15 + 5) : (2 + 3) Wnen entere which follows the order of operations, gives the correct answer Now try problem on your (Part A - Average acceleration3.7 me

meter / second. Note that if you enter the values expression, 15 + 5 and operations into your calculator in exactly the same orde 3 your calculator wll display the wrong answer tor the expressior precedence ol operation errorl An alternative approach in case like this is t0 place parenth numerator and denominator separately in your calculator: (15 + 5) : (2 + 3) Wnen entere which follows the order of operations, gives the correct answer Now try problem on your ( Part A - Average acceleration 3.7 meter /sccond 13.9 meter {stcond Use your calculator t0 evaluale a = 214fecond 7.2 second Vlow Avallable Hint(s} 0.718 mcter/second? 0.718 meter /second? 1.24 meter /sccond? 1135 meter /second? Submit Part B Complete previous parl(s] PartC Complele prevous pur(s]



Answers

Figure $2.19$ shows the acceleration-time graph for a particle in rectilinear motion. The average acceleration in first twenty second is (A) $45 \mathrm{~m} / \mathrm{s}^{2}$ (B) $40 \mathrm{~m} / \mathrm{s}^{2}$ (C) $30 \mathrm{~m} / \mathrm{s}^{2}$ (D) $20 \mathrm{~m} / \mathrm{s}^{2}$

Problem. five. In this problem we have to determine acceleration between three seconds to five seconds. Now from the concept of calculus, we know that acceleration is acceleration is the slope of velocity time curve. So we have to find a slope off a slope of the curve between our time time introvert equals 232 T equals 25 seconds. So this line segment is basically representing the velocity of the car for this time interval. So we have to determine slope of this girl. So far slope. Basically, we need to find the value of theta. So if we drop a perfect, we drop a perpendicular here. So the length of this will be a triangular shape. Let us say this is oh, this is A and this is B. So length of Oh a will be two units since on time access it is between 3-5. Therefore, uh this this length will be two unit and how it will be. Since this oh is corresponding to velocity five and B point is corresponding to 50.0.10. Therefore, this height will be five units. No in this strangle if we draw here again. So this is to to This is five and this is to therefore slow peas represented by Dante to And the 10th it eyes height divided by base. So this will be five divided by two. That is 2.5. Now unit will be meters per second squared. Therefore, acceleration for this time interval acceleration will be 2.5 m per second squared from the given options option B is the correct option.

Okay, so we have a car moving in a straight path, and we record the time and its corresponding position. And so, um, those times and corresponding positions are as follows. So we have the time in seconds and the position in feet. And so a time zero where a position zero at time to were at position 10 feet. And then time four, we're 14 6 20 eight, 30 and 10 were at 36. Okay. And so what we want to do is for the following, we want to find wth e average velocities, um, for these different time frames, and so we want from 0 to 2 seconds. Okay, so the 1st 1 is from 0 to 2 seconds, and so we know that is gonna be, um, 10 minus zero. So it's gonna be the differences of our positions divided by the difference of our time. So this is going to be five, um, feet per second. Okay. And then we want to do part B. We want to do from two to four seconds, and so we would have 14 minus 10 over, um, four minus two. So this would be two feet per second okay. And then part see, we would want to do from for two six seconds. And so 20 minus 14. Over too is three feet per second. And in part D is six to eight seconds. And so we have 30 minus 20 divided by two. Which gives me the five feet per second. Okay. And then what we want to do in E There's a couple things we want to do. An E. We want to estimate the instantaneous velocity at T equal to four seconds, and we want to do it several different ways. We want to. Actually, first way is to determine it by, um, finding the average velocity between two and six seconds. So when we find the average velocity from 2 to 6, so we want to find it from finding the average velocity, um, from two, 26 seconds. And so that is going to be 20 minus 10 divided by four seconds, which is two and 1/2 feet per second. And then what we wanted to the second time is to average my answers from part B, and I believe it waas part. Um see. And so we got, um, that would be two plus three. Divide by two, and we get 2.5 feet per second. Okay, um and then what we want to do is an f We want to do the exact same thing. We want to know the instantaneous velocity, um, at T equal to six seconds, and we want to do it the exact two ways we want to find theat bridge velocity. Um, from I believe it's gonna be from average velocity between, um too. And I've got what we did. 30 and 14 um, four and eight. And so this would be 30 minus 14. Divide by four. So this gives me four feet per second. And then what we want to dio is to, um, find it by, um, averaging my answers from and we're going to do from 4 to 6 from C and D. And so when we do that, we get, um, three plus five, divide by two. Which gives me four feet per second. Okay, Now, in G G says we notice So if you didn't notice, um, when we go back here, um, we got two and 1/2 feet per second. Two and 1/2 feet per second, and then we got four feet per second and four feet per second. So, um, whether we did it for, um, the average or we did it by averaging two separate answers. We got the exact same answers. And so the question is, Is, um well, this always be the case. Um and, um, he's kind of Ah, double edge. Um answer. Um, yes, as long as your increments or the same. Um, And what I mean by that is, if you notice here on both of these, we went from 4 to 6 and 6 to 8. Um, and so six was right exactly in the middle of foreign ate here we went from, um um we went from 2 to 4 and 4 to 6. So, um, if those incremental steps are the same, Yes. Um, but if I try to, um, let's say I wanted to know what it waas at at, um t equal to six seconds. Okay? And what I did waas I found, um first I found from to two six. Then I did six toe eight, and then I average those together. And so if I do 2 to 6, um, this would be 20 minus 10. And I have, um, four seconds here. So that gives me, um, five over to which is 2.5 feet, um, per second. And then from 6 to 8, I have, um, 10 over. Could I do 30 minus 20? Um, and that would be over to which gives me two and 1/2 feet. Who second. So then, if I average those two, if I average those two to get right at six, if I average those two, I would have on five over to which is 2.5 feet per second. Okay, but if I did from, um, two toe eight seconds. Um, right, because we're doing the average. And then we did the average, and then we did 48. Give me that. Um, then 2 to 8 seconds, we would get 30 minus 10 over six, which would be 20 over six. Um, which would give me 10 over three, which is, um, 3.333 feet per second. So, um, so you have to be careful. Um, so, yes, as long as my increments are the same. And that's that time you're looking for is right there in the middle. But if you don't, the answer's gonna be no

So here for part A, we can say that the average acceleration for each interval would be calculated by the average acceleration equaling Delta V divided by Delta T on. And we can say that, uh, and this is going to be taken to be the acceleration at the midpoint of any time interval. So we can say in the spreadsheet, we can say that a to the end plus 1/2 would be equal to B to the n plus one minus v to the end, divided by t to the n plus one minus t to the end. And so oh, um, we can Ah, the second workbook has the accelerations. However, this would be the the except the formula that you were going to use for part a four part B. However, we need the position at the end of each interval and this could be equal. This would be act to the end, plus one to the next interval. Bess essentially would be equal to the initial position at the interval, plus 1/2 times V to the end plus V to the N plus one to the next velocity in the ER data set and then t to the end plus one again minus teach the end. And this we can also represent this as X would be equal to X initial, plus the average velocity times out to tea. However, you want to use this equation because this is much obviously much more specific and tabulated. It would look like this. So these would be the time with the average with the velocity and kilometers per hour and the velocity and meters per second again, we know that one meet one meters per second is gonna be 3.6 kilometers per hour and we know here it would be thine seconds. So this would be, um, for A and these would be for B and then for part C, they wanted to use to grasp the um they want you to go out the acceleration versus time and the distance versus time. And these are what the grafts would look like. You could simply put the data set into excel and then obtain a graph that way. That is the end of the solution. Thank you for watching

So here we have a velocity time graph. Their time is on X axis and velocity is on y axis. And it has said this is a function of the velocity time graph. And we have to find the area bounded by the function the X axis and the time in trouble. So the X axis is the time axis. And time travel is let us say do you want to you too? So this is the area bounded by the function. So this area can be found by using integration. So the area is equal to V duty. We have visa velocity function And time varies from T. 1 to T two. So we had the unit of meter per second and DT is the time which has a unit of seconds. So meter per second into second gives meted and so the area of velocity time graph will be displacement. Hence the correct option is b displacement. We can also see that velocity. We are integrating velocity times time and velocity multiplied by times gifts displacement. So by that argument also the correct option will be displacement. This completes the solution. Thank you


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