An object is thrown straight up. Its height is given by y(t) = 100t − 16t2

Question

An object is thrown straight up. Its height is given by

y(t) = 100t − 16t2

with y in feet and t in seconds. Write a formula for the average velocity on the interval

[2, 2 + h].

Enter your answer as an algebraic expression. You do not have to simplify

Δy

Δt = ___________

Answer 1

The average velocity is Δy/Δt

If this interval [2,2+h] is in time, then Δt = h.

To find Δy, calculate y(2+h) - y(2).

y(2+h) = 100*(2+h) - 16*(2+h)² = 200+100*h - 16*(h²+4*h+4) = 200+100*h -16*h²-64*h - 64 =

136 + 36*h - 16*h²

y(2) = 100*2 - 16*(2)² = 200-64 = 136.

So, Δy = (136 + 36*h -16*h²) - 136 = 36*h - 16*h²

The average velocity is Δy/Δt = (36*h - 16*h²)/h = 36 - 16*h ft/s