Question

I need someone to help me work through this

The distance traveled, in meters, of a coin dropped from a tall building is modeled by the equation d(t) = 4.9t2 where d equals the distance traveled at time t seconds and t equals the time in seconds. What does the average rate of change of d(t) from t = 3 to t = 6 represent?

Answer 1

Answer choices:

The coin travels an average distance of 44.1 meters from 3 seconds to 6 seconds.

The coin falls down with an average speed of 14.7 meters per second from 3 seconds to 6 seconds.

The coin falls down with an average speed of 44.1 meters per second from 3 seconds to 6 seconds.

The coin travels an average distance of 14.7 meters from 3 seconds to 6 seconds

Answer 2

$$
\large{
\text{Average}=\cfrac{1}{6-3}\int_3^6{4.9t^2dt}=\cfrac{1}{3}\times\cfrac{4.9t^3}{3}|^6_3\\
=\cfrac{4.9}{9}\times\left(6^3-3^3\right)\
}
$$

The average rate of change is the average velocity:

$$
\large{
\text{Average Velocity }=\cfrac{1}{6-3}\int_3^6{\cfrac{d}{dt}4.9t^2dt}\\=
\cfrac{1}{3}4.9t^2|_3^6=4.9\times\cfrac{6^2-3^2}{3}=44.1~m/s
}
$$
So it travels at a speed of \(44.1~m/s\) from 3 seconds to 6 seconds

That's it!