# Rates of Change

• March 19th 2009, 06:14 AM
Dickson
Rates of Change
1) s(t) = (t^3) - 8t + 1

Find the displacement during the first 3 seconds.

Do I find s(3) and s(1) and subtract the two? Or what?

2) At what value or values of t does the particle change direction?

Not sure where to start on this one
• March 19th 2009, 06:50 AM
stapel
Quote:

Originally Posted by Dickson
1) s(t) = (t^3) - 8t + 1

Find the displacement during the first 3 seconds.

Do I find s(3) and s(1) and subtract the two?

"The first three seconds" should, I believe, be from t = 0 to t = 3. But "the difference of the ending and starting positions" is the customary definition of "displacement". Check your book (or its worked examples) for specifics.

Quote:

Originally Posted by Dickson
2) At what value or values of t does the particle change direction?

Hint: If the velocity is positive, the particle is moving forward; if negative, then backward. And v(t) = s'(t).

(Wink)
• March 19th 2009, 07:15 AM
Soroban
Hello, Dickson!

Quote:

1) $s(t) \:=\: t^3 - 8t + 1$

Find the displacement during the first 3 seconds.

Do I find s(3) and s(1) and subtract the two?

Yes!

Quote:

2) At what value or values of t does the particle change direction?
Big clue: In order to change the direction, the particle must come to a stop.

The question becomes: When is the velocity equal to zero?

Velocity is the derivative of the position function:
. . . . $v(t) \;=\;s'(t) \;=\;3t^2 - 8 \;=\;0$

We have: . $t^2 \:=\:\tfrac{8}{3} \quad\Rightarrow\quad t \:=\:\pm\sqrt{\tfrac{8}{3}}\quad\Rightarrow\quad t \:=\:\pm\frac{2\sqrt{6}}{3}$