A particle moves along a horizontal line such that its position at time t is specified by s(t) = t^3 - 12t^2 + 36t - 30. What is the total distance travelled by the particle?

Steps to Solve

1. Find the velocity function

To find the velocity of the particle, we need to differentiate the position function $s(t)$.

$$ v(t) = \frac{ds}{dt} = 3t^2 - 24t + 36 $$

2. Determine when the particle changes direction

To find the points where the particle changes direction, set the velocity function equal to zero and solve for $t$:

$$ 3t^2 - 24t + 36 = 0 $$

Dividing the equation by 3 gives:

$$ t^2 - 8t + 12 = 0 $$

Now, factor the quadratic:

$$(t - 6)(t - 2) = 0$$

Thus, the critical points are $t = 2$ and $t = 6$.

3. Evaluate position at critical points and endpoints

We need to evaluate the position function at the critical points and also at $t = 0$ (if specified) to find the positions:

• $s(0) = 0^3 - 12(0)^2 + 36(0) - 30 = -30$
• $s(2) = 2^3 - 12(2)^2 + 36(2) - 30 = 18$
• $s(6) = 6^3 - 12(6)^2 + 36(6) - 30 = -30$

4. Calculate total distance traveled

Now, we calculate the total distance traveled by taking the absolute differences in position between the critical points:

• From $t = 0$ to $t = 2$: $$ |s(2) - s(0)| = |18 - (-30)| = 48$$

• From $t = 2$ to $t = 6$: $$ |s(6) - s(2)| = |-30 - 18| = 48$$

5. Add the distances together

Finally, add together the absolute distances computed:

$$ \text{Total distance} = 48 + 48 = 96$$

Therefore, if the time duration is limited to a specific range, ensure to adjust calculations based on that.