Question

by the function ( h(t)=300 - 16t^{2} ). which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall?
( h(3)-h(0) )
( \frac{h(3)-h(0)}{3} )
( \frac{h(3)}{3} )
( \frac{h(3)-h(0)}{3} )

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is given by \(\frac{f(b)-f(a)}{b - a}\).

Step2: Identify \(a\), \(b\) and the function

Here, the function is \(h(t)=300 - 16t^{2}\), \(a = 0\), \(b=3\).

Step3: Substitute into the formula

Substituting into the average - rate - of - change formula \(\frac{h(3)-h(0)}{3-0}=\frac{h(3)-h(0)}{3}\)

Answer:

\(\frac{h(3)-h(0)}{3}\)