Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Help!!!

OpenStudy (anonymous):

The function \[s=-t^3+6t^2-12t, 0\le3\] gives the position of a body moving on a coordinate line, with s in meters and t in seconds. a. Find the body's displacement and average velocity for the given time interval b. Find the body's speed and acceleration at the endpoints of the interval c. When, if ever, during the interval does the body change direction

OpenStudy (mertsj):

Did you take the first derivative of the distance function which would be the velocity function?

OpenStudy (anonymous):

\[-3t^2+12t-12\]

OpenStudy (mertsj):

a. Find the position at 0 and at 3 and subtract to find the displacement for the interval

OpenStudy (anonymous):

-9?

OpenStudy (mertsj):

That's what I got.

OpenStudy (mertsj):

Find velocity at beginning and end of interval, subtract, divide by 3 to get average velocity

OpenStudy (anonymous):

how?

OpenStudy (mertsj):

What is the velocity function?

OpenStudy (anonymous):

\[-3t^2+12t-12\]

OpenStudy (mertsj):

What is the velocity at t = 0?

OpenStudy (anonymous):

-12

OpenStudy (mertsj):

What is the velocity at t = 3?

OpenStudy (anonymous):

-3

OpenStudy (mertsj):

What is -3minus -9?

OpenStudy (anonymous):

6

OpenStudy (mertsj):

Whoops. What is -3 minus -12 I should have said.

OpenStudy (anonymous):

9

OpenStudy (mertsj):

What is 9 divided by 3?

OpenStudy (anonymous):

3

OpenStudy (mertsj):

ok. average velocity : (velocity at end of interval - velocity at beginning of interval) divided by length of interval

OpenStudy (mertsj):

Now b. You've already found the velocity at the beginning and end of the interval. The acceleration is the second derivative of the distance function which is the first derivative of the velocity function so find the acceleration function.

OpenStudy (anonymous):

-6t+12

OpenStudy (mertsj):

Plug in 0 and 3 to find the acceleration at the beginning and end of the interval

OpenStudy (mertsj):

Did you get 12 and -6?

OpenStudy (anonymous):

yes

OpenStudy (mertsj):

Now,c. When does it change direction. The position function would have a high point or a low point if direction changed. That means the slope of the tangent at that point would be 0. So take the slope function, which is the first derivative, which is the velocity function, and set it equal to 0 and solve to see if there is any t value between 0 and 3 at which the slope is 0

OpenStudy (anonymous):

at 2?

OpenStudy (mertsj):

|dw:1364604883611:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!