I need help with a Calculus problem?

Question

If a stone is thrown down at 150 ft/sec from a height of 1000 feet, its height after t seconds is given by s= 1000 − 150t − 16t^2.

(a) Find its average velocity over the period [1, 2]. I got -198ft/sec

(b) Estimate its instantaneous velocity at time t = 2.
This is the one I'm having trouble with.

Please find the answer, and let me know how you got it.

Whoever gets answer first gets picked for "Best Answer!"

Answer 1

Take the derivative of the position function. Then substitute in t = 2 and that will give you its instantaneous velocity at 2 seconds. Remember, the derivative of the position function with respect to t is the velocity function.

s = 1000 − 150 t − 16 t².
ds = 0 -150 dt - 16 (2 t) dt
ds/dt = 150 - 16 (2 t)
ds/dt = -150 - 16 [2 (2)]
ds/dt = -150 - 64
ds/dt = - 214 ft/sec

So, the velocity of the stone is - 214 ft/sec at t = 2 seconds.

Answer 2

After 2 secs

instant velocity =ds/dt

= -150-32t

so after 2 secs = -214 ft/sec