Question

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.

Answer 1

8.4m/s

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

Answer 2

8.4m/s

Explanation:

the information we are given is:

initial time:
`t_(1)=4s`

final time:
`t_(2)=8s`

thus, the interval of time is:
`t=t_(2)-t_(1)=8s-4s=4s`

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

`d(t)=0.7t^2`

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

At a time of 4 seconds ⇒
`t_(1)=4s`

the distance at that time is:

`d(4)=0.7(4)^2\\d(4)=0.7(16)\\d(4)=11.2m`

also, the distance at a time of 8 seconds ⇒
`t_(2)=8s`

and the distance at this time is:

`d(8)=0.7(8)^2\\d(8)=0.7(64)\\d(8)=44.8m\\`

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

`d=d(8)-d(4)\\d=44.8m-11.2m\\d=33.6m`

and finally we use the formula for the average speed:

`s=(d)/(t)`

where
`d` is the distance traveled in
`t` time:

`s=(33.6m)/(4s) \\s=8.4m/s`

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