Jazelyn1
contestada

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 model by the function d(t) =0.8t^2. what is the average speed, in meters per second, of the rock between 5 and 10 seconds after it was dropped?

Respuesta :

jashae2003
Given
d(t) = 0.8t ^2

t= 5 secs
d(5) = 0.8(5)^2 = 0.8 * 25 = 20
t= 10 secs
d(10) = 0.8(10)^2 = 0.8 * 100 = 80

20 m / 5 secs = 4 meters per sec
80 m / 10 sec = 8 meters per sec

4 mps + 8 mps = 12 mps
12 mps / 2 = 6 mps
6 mps is the average speed of the rock between 5 and 10 sec after being dropped

Answer:

6

Step-by-step explanation:

Given

d(t) = 0.8t ^2

t= 5 secs

d(5) = 0.8(5)^2 = 0.8 * 25 = 20

t= 10 secs

d(10) = 0.8(10)^2 = 0.8 * 100 = 80

20 m / 5 secs = 4 meters per sec

80 m / 10 sec = 8 meters per sec

4 mps + 8 mps = 12 mps

12 mps / 2 = 6 mps

6 mps is the average speed of the rock between 5 and 10 sec after being dropped