A rock is dropped off a ledge on the moon according to the
formula d(t) = .7t^2, find the average speed, in meters per
second, between 4 and 8 seconds after it was dropped.

8.4m/s

Step-by-step explanation:

the information we are given is:

initial time:

final time:

thus, the interval of time is:

according to the statement, the distance at a certain time is given by:

To find the average distance we need to find first the total distance traveled in those 4 seconds .

At a time of 4 seconds ⇒

the distance at that time is:

also, the distance at a time of 8 seconds ⇒

and the distance at this time is:

Now we can find the distance traveled between 4 and 8 seconds:

and finally we use the formula for the average speed:

where is the distance traveled in time:

the average speed in meters per second: 8.4m/s

8.4m/s

Step-by-step explanation:

Speed or velocity is defined as the change in distance of a body with respect to time. It is expressed as:

Speed = distance/time

Given distance d(t) = 0.7t²

When t = 4secs

d4) = 0.7 × 4²

d(4) = 11.2m

At t = 8secs

d(8) = 0.7× 8²

d(8) = 44.8m

Since speed = distance/time

Speed at t = 4secs = 11.2/4

= 2.8m/s

Speed at t = 8secs = 44.8/8

= 5.6m/s

average speed, in meters per seconds between 4 and 8 seconds after it was dropped will be ∆d/∆t

= 44.8-11.2/8-4

= 33.6/4

= 8.4m/s

step-by-step explanation:

your answer is 125y5