anonymous
  • anonymous
In order to find the instantaneous velocity of this equation: x = 34 + 10t-2t^3, between 0 secs and 3.0 secs, I would need to take the derivative of the equation, plug in my t values from 0 to 3.0 secs and divide by 3.0 secs right?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
The question I'm trying to figure out also asked for the average velocity and I got -8.0m/s
anonymous
  • anonymous
It's not the "instantaneous velocity" if there's a time range. But yes, to find the instantaneous velocity at any given time you take the derivative, which will be: \[x'=v=10-6t ^{2}\] Then plug in the time you desire for t, and voila.
anonymous
  • anonymous
For average velocity, between times \[t=0, t=3\] Will be the total displacement between the two times over the time, a la \[\Delta x / \Delta t\] So take the original equation for x, evaluate it at t=3, then evaluate it at t=0. Subtract and divide by t=3.

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anonymous
  • anonymous
I get that but its also asking me for the when between 0 and 3.0 where the instantaneous velocity reaches zero. The time intervals it gives me is 0.5,1.0,1.5,2.0,2.5, and 3.0.
anonymous
  • anonymous
I have the derivative of the function but when I plug in each t value and divide that by time i oroginally plugged into the eqaution none of them seem to equal zero.

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