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
"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.
Hint: If the velocity is positive, the particle is moving forward; if negative, then backward. And v(t) = s'(t).
Hello, Dickson!
Yes!1) $\displaystyle s(t) \:=\: t^3 - 8t + 1$
Find the displacement during the first 3 seconds.
Do I find s(3) and s(1) and subtract the two?
Big clue: In order to change the direction, the particle must come to a stop.2) At what value or values of t does the particle change direction?
The question becomes: When is the velocity equal to zero?
Velocity is the derivative of the position function:
. . . . $\displaystyle v(t) \;=\;s'(t) \;=\;3t^2 - 8 \;=\;0$
We have: .$\displaystyle t^2 \:=\:\tfrac{8}{3} \quad\Rightarrow\quad t \:=\:\pm\sqrt{\tfrac{8}{3}}\quad\Rightarrow\quad t \:=\:\pm\frac{2\sqrt{6}}{3} $