Difficulty with a basic motion problem

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To solve the motion problem defined by the position function x = 2.30t^5, the average velocity between t = 1.0 s and t = 2.0 s can be calculated using the formula for average velocity, which is (x(t_b) - x(t_a)) / (t_b - t_a). The average acceleration is similarly derived from the change in velocity over the time interval. For instantaneous values, the velocity is found by taking the derivative of the position function, while the acceleration is the derivative of the velocity function. Clarification on calculating the two velocities for average acceleration is provided, emphasizing the need for derivatives of the position function.
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This time I'm having trouble with this problem:

If the position of an object is given by x = 2.30t^5, where x is measured in meters and t in seconds, find (a) the average velocity and (b) the average acceleration between t = 1.0 s and t = 2.0 s. Then find (c) the instantaneous velocity v and (d) the instantaneous acceleration a at t = 1.0 s. Next find (e) v and (f) a at t = 2.0 s.

I have solved a, but am uncertain how to obtain the two velocities needed for b according to the formula for average velocity. What do I need to do?
 
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the velocity function is the time-derivitive of the location function.

the acceleration function is the time-derivitive of the velocity function.
 
the average velocity forumula is [f(b)-f(a)]/b-a, the avg. acceleration formula is essentially the same except that for the values of f(b) and f(a) you need to take the derrivative of the position equation to give the velocity values
 
aud11888 means that average velocity = \frac{x(t_b) - x(t_a)} {t_b - t_a}
 
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