The height, 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?

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.

For t = 3 we get:

d(3) = 4.9*3^2 = 4.9*9 = 44.1

For t=6 we get:

d(6) = 4,9*6^2 = 176.4

The difference of these 2 values gets us distance traveled between t=3 and t=6 seconds.

d(6) - d(3) = 132.3 meters

if we divide this number with number of seconds that it takes coin to cover that distence we get:
132.3/(6-3) = 44.1
which is average speed between t=3 and t=6

answer is last option.

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tardymanchester tardymanchester · Answer 2 · 2019-01-14T07:52:26+01:00

Answer:

Option C

Step-by-step explanation:

Given : A coin dropped from a tall building is modeled by the equation [tex]d(t) = 4.9t^2[/tex] where, d is the distance traveled at time t seconds and t equals the time in seconds.

To find: The average rate of change of d(t) from t = 3 to t = 6

[tex]d(t) = 4.9t^2[/tex]

For t = 3 :

[tex]d(3) = 4.9\times3^2 = 4.9\times9 = 44.1[/tex]

For t=6 :

[tex]d(6) = 4.9\times6^2 =4.9\times36= 176.4[/tex]

The difference of distance traveled between t=3 and t=6 seconds.

[tex]d(6) - d(3) = 176.4-44.1= 132.3[/tex]

Now, we find the average speed :

[tex]Speed=\frac{Distance}{Time}[/tex]

[tex]Speed=\frac{132.3}{6-3}[/tex]

[tex]Speed=\frac{132.3}{3}[/tex]

[tex]Speed=44.1m[/tex]

Therefore, Option C - The coin falls down with an average speed of 44.1 meter from 3 seconds to 6 seconds.