Question

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the function h(t)=300-16t squared which expression could be used to determine the average rate at which the object falls during the first 3 seconds of its fall

Answer 1

The average rate at which the object falls during the first 3 seconds of its fall is -144 feet per second.

Step-by-step explanation:

The average rate at which the object falls during the first 3 seconds of its fall can be determined by calculating the change in height of the object over the change in time. In this case, the height function is given by h(t) = 300 - 16t^2. To find the average rate of change, we can calculate the difference in height between t=0 and t=3 and divide it by the difference in time, which is 3 seconds.

So, h(3) - h(0) = (300 - 16(3^2)) - (300 - 16(0^2)) = (300 - 144) - (300 - 0) = 156 - 300 = -144.

The negative sign indicates that the object is falling. Therefore, the average rate at which the object falls during the first 3 seconds of its fall is -144 feet per second.

Answer 2

`\bf slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby \begin{array}{llll} average\ rate\\ of\ change \end{array}\\\\ -------------------------------\\\\ f(x)= 300-16t \qquad \begin{cases} x_1=0\\ x_2=3 \end{cases}\implies \cfrac{f(3)-f(0)}{3-0} \\\\\\ \cfrac{[300-16(3)]~~-~~[300-16(0)]}{3}\implies \cfrac{[300-48]~-~[300]}{3} \\\\\\ \cfrac{252-300}{3}\implies -\cfrac{48}{3}\implies -\cfrac{16}{1}\implies -16`

is a negative rate, because the object is falling down, at 16 feet every passing second.