The height of a falling object (subject to air resistance) is given by h(t) = 400 − 39t − 157e−t/4

Question

The height of a falling object (subject to air resistance) is given by

h(t) = 400 − 39t − 157e^−t/4

with h in meters and t in seconds. Write a formula for the average velocity on the interval

[2, a].

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

Δh

Δt = _______

Answer 1

The key is that the average velocity is defined as integral from t₁ to t₂ of vdt divided by the (t₂-t₁)

The integral of vdt = integral of (dh/dt)dt which yields h(t₂)-h(t₁) . The equation for the average velocity is

(h(t₂)-h(t₁))/(t₂-t₁) which is why your are calculating Δh/Δt

You can do this without simplifying. Just plug in what you get for h(2)-h(a))/(2-a)