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?
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The question I'm trying to figure out also asked for the average velocity and I got -8.0m/s
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:
Then plug in the time you desire for t, and voila.
For average velocity, between times
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.