Question

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

Average Rate of change of d(t) is 11.43

Explanation:

Given:

d(t) = 4.9 t²

where d(t) gives distance covered by coin in time t.

To find: The Average Rate of change of d(t) from t = 3 to t = 6.

We know that Average rate of change, A of f(x) =
`(f(a)-f(b))/(a-b)`

here, a= 6 and b = 3

So, Average rate of change =
`(d(6)-d(3))/(6-3)`

=
`(4.9(6)^2-4.9(3)^2)/(3)`

=
`(4.9*16-4.9*9)/(3)`

=
`(78.4-44.1)/(3)`

=
`(34.3)/(3)`

=
`11.43`

Therefore, Average Rate of change of d(t) is 11.43

Answer 2

The rate of change of the distance traveled by the coin is obtained by differentiating the function d(t) and substituting the value of t. The average rate of change in d(t) from t = 3 to t = 6 represent the distance traveled by the coin between those time intervals.