# Difficulty with a basic motion problem

1. Jan 23, 2006

### frankfjf

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?

2. Jan 23, 2006

### lightgrav

the velocity function is the time-derivitive of the location function.

the acceleration function is the time-derivitive of the velocity function.

3. Jan 23, 2006

### aud11888

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

4. Jan 23, 2006

### lightgrav

aud11888 means that average velocity = $$\frac{x(t_b) - x(t_a)} {t_b - t_a}$$

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