One more thing with motion problem

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I'm referring to the same problem as before: https://www.physicsforums.com/showthread.php?t=107835

But now my question is how do I figure out parts c and e? My textbook says to just take the slope of the position-time curve. But if t = 1 in part c, how do I setup the slope? It'll just be something over 1, or that something. My problem is if I take the slope of the change in position over 1, I just get my answer to part a again, the average velocity or 71.3. I'm stuck. Please help?
 
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the instantaneous velocity function is given by the time-derivative of the position function.

do you have an expression for the velocity at some time t?
 
eep said:
the instantaneous velocity function is given by the time-derivative of the position function.

So does that mean that I take the derivative of x = 2.30^5 and plug in 1s and 2s, respectively for parts c and e?
 
yup! and the function for instantaneous acceleration is the time-derivative of the velocity function. One could also take the slope of the position function at t=1.0, since that's what the formula for the instantaneous velocity is giving you.
 
Thanks! So basically..

For the instantaneous velocity take the first derivative, and for the instantaneous acceleration take the second derivative? (Of x = 2.30t^5)
 
Correct. Of *any* position function.