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Velocity, Position, and MVT?

Velocity, Position, and MVT? Topic: How to write a polynomial expression
June 26, 2019 / By Flora
Question: My Calculus teacher and I are having a disagreement. The question reads: A particle moves along the x-axis so that is velocity at any time greater than 0 is given by v(t) = 3t^2 -2t -1. he position x(t) is 5 for t = 2. a. Write a polynomial expression for the position of the particle for any time greater than 0. If you work that out, you get x(t) = t^3 - t^2 - t + 3 The disagreement is when we get to part b, which is: b. For what values of t, 0 is less than or equal to t is less than or equal to 3, is the particle's instantaneous velocity the same as its average velocity on the closed interval [0,3]. My teacher and I don't agree on how to do this problem, so of course we have different answers. What does Yahoo! answers say the answer is?
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Best Answers: Velocity, Position, and MVT?

Dalia Dalia | 1 day ago
I'm guessing that the disagreement arises over the distinction between speed and velocity, and the fact that the direction of the velocity changes when t=1. Average velocity = (change in position)/(time interval) Now x(0) = 3 and x(3) = 18 Hence average velocity = (18-3) / 3 = 5 units in the positive direction We require to solve 3t² - 2t - 1 = 5 3t² - 2t - 6 = 0 t = [2 ±√76]/6 .. = [1 + √19] / 3 since we need the positive root only. It's roughly 1.8. Average speed = (distance covered) / (time taken) x(1) = 2 Hence distance covered between t=0 and t=1 is 3 - 2 = 1 unit in the negative direction Distance covered between t=1 and t=3 is 18 - 2 = 16 unit in the positive direction. Total distance covered = 16 + 1 = 17 unit Therefore average speed = 17/3 = 5.66666..... unit in the positive direction Now this was not the question, but if it had been to find when the instantaneous speed is equal to the average speed, then we would need to solve 3t² - 2t - 1 = 17/3 9t² - 6t - 20 = 0 t = [6 ±√756] / 18 and the positive root is roughly 1.9
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Dalia Originally Answered: Should I leave my new position to go work for the person that hired me into my current position?
I was in retail for 14 years, 2 in sales, 10 loss prevention and 2 in AP and AR. I received a call from an ex-L.P co worker. There was a position where he worked in a clothing warehouse (MORE RETAIL)for front desk receptionist, then did Inventory Control. It was further away but I did make more money. I was very happy there for four years and moved on. I also made the right choice (for me). I learned alot from my co worker and got a job as an Import Corprate Assistant with a global company. No more retail for me. :) Good Luck to you !
Dalia Originally Answered: Should I leave my new position to go work for the person that hired me into my current position?
Hi Shan already your company has shown you what they do with employee's , they told her they no longer needed her and puff just like that she's toast. So if they can do it to her what makes you so special. So what 5 bucks its cheaper to pay you the 5 bucks than a yearly salary that she was at and has added extra work for you. there buying you at a cheap price. If you have a good rapport with this woman ( and only you know how you felt working with her) I'd say go to where she's at. way the pro's and cons how is the company treating you? But if there that easy to get her outta there saying her job is redundant what does that say about the company its all about figures and money. DO you trust this woman? was she good to youl. Do you see potential with her, how is here work ethics. If you have more pro's than cons I'd go with her. Your company has already shown you what there about deep down. Was she a good work to be fair to the company did she have issues did she just sail along on other peoples back or a free loader never doing more than she needed to if you can size her up and know she was top notch and have no doubts then i would go with her.. What would I do . I"l d look at her history. I'd see how she was as a director or manager, YOu can tell in meetings with her if she was any good or if back stabbing the company or slacked off. I would disect her position and look back to see how she was in her job if she was a straight shooter up front and hard working and had the employee's back etc... if it all points good to her I would go. I would also way how the company has treated me and how they have treated other individuals that says alot. Then make your decision don't jump ship until you 've done your homework.
Dalia Originally Answered: Should I leave my new position to go work for the person that hired me into my current position?
Make lists of the pros and cons of current job and of the one you've been offered. If your CNA license isn't being supported at current one, how important is that to your future plans? What about transportation costs?(remember that if you drive, your car might need to be replaced sooner ) What if your car is in the shop? How will you get there? Do you have to consider driving in the snow or other bad weather? That would be an extra hour a day of driving, even without bad weather or traffic delays. Are there health benefits at either job? Any chance you would want to move in order to be near new job? Decisions like this are difficult. Best wishes to you.

Blanda Blanda
Velocity Function: v(t) = 3t² - 2t -1 a.) The Position Function, x(t), is the integral of the Velocity Function: x(t) = ∫ [v(t)] dt x(t) = ∫ [3t² - 2t - 1] dt x(t) = (6t³ / 3) - (2t² / 2) - t + c x(t) = 2t³ - t² - t + c Given: x(t) = 5 for t = 2, or x(2) = 5, so 2(2)³ - 2(2)² - 2(2) + c = 5 2(8) - 2(4) - 4 + c = 5 16 - 8 - 4 + c = 5 4 + c = 5 c = 5 - 4 c = 1, so x(t) = 2t³ - 2t² - 2t + 1 b.) t1 = 0 t2 = 3 x(0) = 2(0)³ - (0)² - 0 + 1 x(0) = 1 unit x(3) = 2(3)³ - 2(3)² - 2(3) + 1 x(3) = 2(27) - 2(9) - 2(3) + 1 x(3) = 54 - 18 - 6 + 1 x(3) = 31 units v(0) = 3(0)² - 2(0) - 1 v(0) = - 1 unit/sec. v(3) = 3(3)² - 2(3) - 1 v(3) = 3(9) - 6 - 1 v(3) = 27 - 6 - 1 v(3) = 20 units/sec. Average Velocity, va = [x(3) - x(0)] / [v(3) - v(3)] va = (31 - 1) / [20 - (- 1)] va = 30 / (20 + 1) va = 30 / 21 va = 1.43 units/secs. Instantaneous Velocity = Instantaneous Velocity, so v(t) = va, or 3t² - 2t -1 = 1.43 3t² - 2t = 1.43 + 1 3t² - 2t = 2.43 3(t² - 2/3 t) = 2.43 t² - 2/3 t = 2.43 / 3 t² - 2/3 t = 0.81 t² - 0.667t = 0.81 t² - 0.667 + 0.111 = 0.81 + 0.111 (t - 0.334)² = 0.921 t - 0.334 = √0.921 t - 0.334 = ± 0.96 t = ± 0.96 + 0.334 If t = 0.96 + 0.334, t = 1.29 If t = - 0.96 + 0.334, t = - 0.626 Given: 0 ≤ t ≤ 3 , so t = 1.29 Average velocity and instantaneous velocity are the same at t = 1.29 secs. .....................
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Blanda Originally Answered: Acceleration and velocity?
Christian - Please, let's start with "speed". "Speed" is how fast you are moving. I can tell you that I am driving at a speed of 30 mph... 44 feet per second... on East Elm Street. You don't know where I am. You don't know where I'm going. If you live on East Elm Street you don't know whether you will see me or not. Now I'll add that I am going north at 44 ft/s on East Elm St. NOW you have my velocity: Velocity is the combination of the speed I just talked about PLUS the direction I'm moving: I'm going north on East Elm St. at a speed of 30 mph. My velocity is 30mph, North. Velocity is speed and direction. If those considerations of speed and velocity make sense to you, I believe that acceleration will be very easy: Acceleration is any change in velocity. That means any change in speed, any change in direction. When I stop for the light at Physics Blvd. I have to slow from 44 ft/s to zero ft/s. It took 5 1/2 seconds. I had an acceleration of 44/5.5 = 8 ft/sec/sec. So, to your specific questions: Your first line was correct for speed. Your second line (Velocity is...) is incorrect - but I think we've addressed that Your third line is perfect.. Your second paragraph (two sentences) is correct... but you are very tentative. I'm interpreting what you wrote as saying that you thought "acceleration" only applied to change in speed. We're passed that... I hope. By definition you will accelerate... you will change your velocity when you move into the left lane from the right: You changed your direction during the lane switch. (Now, dam*it, either pass someone - or get back in the right lane!) 'We good?, Richard
Blanda Originally Answered: Acceleration and velocity?
Your definition of velocity is incorrect. Velocity is the displacement (not distance) divided by time. It is a vector quantity (has magnitude and direction) To understand velocity better, let's talk about your perfect example of driving a car. If you step on gas, you move faster and faster - the magnitude of the velocity changes, therefore you accelerate. If you drive at constant speed and change the direction, the velocity changes as well (the speed does not change), therefore you accelerate.

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