Question

an object is falling such that the distance it travels is given as a function by the formula d(t)=16t².
(a) fill out the following table for the distance the object has traveled each half second (using calculus)
(b) how much faster, in feet per second, is the object falling over the last two seconds than the first two seconds? show your analysis.
(c) what is the objects average speed for the entire four seconds?

Explanation:

Step1: Recall the formula for average speed

The formula for average speed is \(v_{avg}=\frac{\Delta s}{\Delta t}\), where \(\Delta s\) is the total distance traveled and \(\Delta t\) is the total time elapsed.

Step2: Find the total distance traveled

We are given the function \(d(t) = 16t^{2}\). For \(t = 4\), \(d(4)=16\times4^{2}=16\times16 = 256\) feet.

Step3: Calculate the average speed

Since \(\Delta t=4\) seconds and \(\Delta s = 256\) feet, using the formula \(v_{avg}=\frac{\Delta s}{\Delta t}\), we substitute \(\Delta s = 256\) and \(\Delta t = 4\). So \(v_{avg}=\frac{256}{4}\)

Answer:

\(64\) feet per second