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Topic: **How to write a polynomial expression****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?

June 26, 2019 / By Flora

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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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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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

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.

If you have your own answer to the question how to write a polynomial expression, then you can write your own version, using the form below for an extended answer.