Respuesta :
your average speed is the average difference over time (in this case 3 seconds)
[tex] \frac{h(3)-h(0)}{3}=\\ \frac{300-16*3^2-(300-16*0^2)}{3}=\\ \frac{300-16*3^2-(300)}{3}=\\ \frac{-16*3^2}{3}=\\ -16*3=-48[/tex]
[tex] \frac{h(3)-h(0)}{3}=\\ \frac{300-16*3^2-(300-16*0^2)}{3}=\\ \frac{300-16*3^2-(300)}{3}=\\ \frac{-16*3^2}{3}=\\ -16*3=-48[/tex]
Answer:
-96 feet/sec.
Step-by-step explanation:
The height h of a falling object after t second when it is dropped from a platform 300 feet above the ground is modeled by the function h (t) = 300 - 16t²
Then speed or average rate by which the object falls from height h will be [tex]\frac{d[h(t)]}{dt}=\frac{d(300-16t)^{2} }{dt}[/tex]
Speed = [tex]\frac{d}{dt}(300)[/tex] - [tex]\frac{d}{dt}(16t^{2} )[/tex]
= -16 (2t)
Speed = -32t
After 3 second speed of object will be
Speed = -32(3)
= -96 feet/second
Therefore, the average rate will be -96 feet/sec. during the first 3 seconds of its fall.