Question

An astronaut drops a rock off the edge of a cliff on the moon. The distance d(t), in meters, the rock travels after t seconds can modeled by the function d(t)= .5x^2. What is the average speed, in meters per second, of the rock between 5 and 10 seconds after it was dropped

Answer 1

The average speed is 7.5
`( m)/( s)`

Explanation:

Speed ​​is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it.

Then, the average speed relates the change in position to the time taken to effect that change:

`speed=(change in position)/(change in time)`

The change in position is the difference between the final position and the initial position, while the change in time is the difference between the final and initial time.

In this case, the starting and ending positions are calculated by replacing the times in the function of the distance d (t):

d(5)= 0.5*5²= 12.5 m

d(10)= 0.5*10²= 50 m

So:

• change in position= final position - initial position= 50 m - 12.5 m= 37.5 m
• change in time= 10 s - 5 s= 5 s

Replacing in the definition of speed:

`speed=(37.5 m)/(5 s)`

and solving you get:

speed= 7.5
`( m)/( s)`

The average speed is 7.5
`( m)/( s)`