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# 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?

## Best Answers: Velocity, Position, and MVT?

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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