# Homework Help: One more thing with motion problem

1. Jan 23, 2006

### frankfjf

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?

2. Jan 23, 2006

### eep

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?

3. Jan 23, 2006

### frankfjf

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?

4. Jan 23, 2006

### eep

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.

5. Jan 23, 2006

### frankfjf

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)

6. Jan 23, 2006

### eep

Correct. Of *any* position function.