Question

an astronaut drops a rock into a crater on the moon. The distance, d(t), in meters, the rock travels after t seconds can be modeled by the function d(t) =0.8t2. what is the average speed, in meters per second, of the rock between 5 and 10 seconds after it was dropped?

Answer 1

6 m / s

Explanation:

We know that speed is distance over time, so what we will do is calculate the distance between 5 and 10 seconds, and then we calculate each speed and then average.

d (5) = 0.8 * 5 ^ 2 = 20

v (5) = 20/5 = 4

d (6) = 0.8 * 6 ^ 2 = 28.8

v (6) = 28.8 / 6 = 4.8

d (7) = 0.8 * 7 ^ 2 = 39.2

v (7) = 39.2 / 7 = 5.6

d (8) = 0.8 * 8 ^ 2 = 51.2

v (8) = 51.2 / 8 = 6.4

d (9) = 0.8 * 9 ^ 2 = 64.8

v (9) = 64.8 / 9 = 7.2

d (10) = 0.8 * 10 ^ 2 = 80

v (10) = 80/10 = 8

If we average we have:

(4 + 4.8 + 5.6 + 6.4 + 7.2 + 8) / 6 = 6

In other words, the average speed is 6 m / s