A point moves along the x-axis, and its position, x, at time t seconds is given by p(t) = t^3 – 6t^2 + 5. Find the total distance the point travels from t = – 1 to t = 6.

2. Originally Posted by bobby77
A point moves along the x-axis, and its position, x, at time t seconds is given by p(t) = t^3 – 6t^2 + 5. Find the total distance the point travels from t = – 1 to t = 6.
To put it in a single statement, what you are looking for is:
$s=\int_{-1}^6 dt \, |v(t)|$

In other words, we need to find out if and where the object turns around.

$v(t)=3t^2-12t$

The velocity has zeros at t = 0 s and t= 4 s. Since v(t) is positive at t = -1, the object starts moving from position -2 and moves in the +x direction to t = 0 s at position 5 (7 units total distance) and turns around. It goes in the -x direction to t = 4 s at position -27 (34 units total distance) and turns around. It goes in the +x direction to t = 6 s at position 5 (66 units total distance.) So the object moves a total of 66 units.

-Dan

3. Originally Posted by topsquark
To put it in a single statement, what you are looking for is:
$s=\int_{-1}^6 dt \, |v(t)|$

In other words, we need to find out if and where the object turns around.

$v(t)=3t^2-12t$

The velocity has zeros at t = 0 s and t= 4 s. Since v(t) is positive at t = -1, the object starts moving from position -2 and moves in the +x direction to t = 0 s at position 5 (7 units total distance) and turns around. It goes in the -x direction to t = 4 s at position -27 (34 units total distance) and turns around. It goes in the +x direction to t = 6 s at position 5 (66 units total distance.) So the object moves a total of 66 units.

-Dan
Topsquark you learned that trick using absolute value from me

4. What? Did you trademark it when I wasn't looking?

To all who post here: I learned the previous integral form from ThePerfectHacker!

(I promise I'll pay you a tithe from any moneys I get from answering bobby77's question. Would you prefer cash or credit? )

-Dan