Calculus problem I need help with please show your work

A ball is dropped from a state of rest at time t=0.

The distance traveled after t seconds is s(t)=16t^2 ft.

Compute the AVERAGE VELOCITY over the time intervals [3,3.01], [3,3.005], [3,3.001], and [3,3.0005].

Use this computation to estimate the ball's INSTANTANEOUS VELOCITY at t=3.

Compare this velocity to s'(3).

I have already done the work I just need to check to see if its right. I would ask my boyfriend, but he's at work right now.

please explain the process as you work through the problem in case I didn't do something right.

For the first portion of this problem, you have to use the difference quotient to solve for the average velocity:

V_avg = [S(t + h) - S(t) ]/h

For time intervals [3, 3.01]:

V_avg = [16(3.01)^2 - 16(3.0)^2]/0.01
=96.16

For time intervals [3, 3.005]:

V_avg = [16(3.005)^2 - 16(3.0)^2]/0.005
=96.08

For time intervals [3, 3.001]:

V_avg = [16(3.001)^2 - 16(3.0)^2]/0.001
=96.016

For time intervals [3, 3.0005]:

V_avg = [16(3.0005)^2 - 16(3.0)^2]/0.0005
=96.008

Now, taking the derivative of the distance function over time:
s(t) = 16(t^2)
s'(t) = 32t
s'(3) = 96 <-----this is the instantaneous velocity as compared to the average velocities. As you can recall, the derivative is simply the limit as h ---> infinity of the difference quotient.

I am sure your calculations are good, but just in case ...

S(t) = 16t^2 ft.

a) S(3) = 16(3)^2 = 144
S(3.01) = 16(3.01)^2 = 144.9616
Average velocity = 0.9616/0.01 = 96.16

b) S(3) = 16(3)^2 = 144
S(3.005) = 16(3.005)^2 = 144.4804
Average velocity = 0.4804/0.005 = 96.08

c) S(3) = 16(3)^2 = 144
S(3.001) = 16(3.001)^2 = 144.096016
Average velocity = 0.096016/0.001 = 96.016

You do d)

The estimate of ball's INSTANTANEOUS VELOCITY at t = 3 is about 96 ft/sec
S'(3) = dS/dt at 3
dS/dt(3) = 32t at t = 3 = 96 ft/sec

Regards - Ian