To address the question shown in your image, the key distinction is between displacement and distance traveled:

1. Displacement is the integral of the velocity function over the given time interval, which accounts for the direction of motion (negative velocity reduces the total).
2. Distance traveled is the integral of the absolute value of the velocity function over the given interval, meaning it accounts for all motion as positive regardless of direction.

Steps to Solve:

Given:

$v(t) = t^2 - 2t - 8$ on $[2, 5]$.

(a) Displacement:

To calculate the displacement: $\text{Displacement} = \int_{2}^{5} v(t) \, dt$ This is already computed as 60 meters, as shown in the image.

(b) Distance Traveled:

We need to account for changes in direction by finding where $v(t) = 0$ (velocity changes sign).

1. Solve $v(t) = 0$: $t^2 - 2t - 8 = 0$ Factoring: $(t - 4)(t + 2) = 0$ So, $t = 4$ (within the interval $[2, 5]$) and $t = -2$ (not in the interval).

2. Split the interval at $t = 4$: The motion direction changes at $t = 4$. Therefore, evaluate the absolute value of $v(t)$ on $[2, 4]$ and $[4, 5]$.

3. Integrate $|v(t)|$:

   • On $[2, 4]$, $v(t) < 0$, so use $-v(t)$: $\int_{2}^{4} |v(t)| \, dt = \int_{2}^{4} -(t^2 - 2t - 8) \, dt$
   • On $[4, 5]$, $v(t) > 0$, so use $v(t)$: $\int_{4}^{5} |v(t)| \, dt = \int_{4}^{5} (t^2 - 2t - 8) \, dt$

4. Compute the integrals:

   For $[2, 4]$: $\int_{2}^{4} -(t^2 - 2t - 8) \, dt = \int_{2}^{4} (-t^2 + 2t + 8) \, dt$ Compute the antiderivative: $\int (-t^2 + 2t + 8) \, dt = -\frac{t^3}{3} + t^2 + 8t$ Evaluate from 2 to 4: [ \left[-\frac{4^3}{3} + 4^2 + 8(4)\right] - \left[-\frac{2^3}{3} + 2^2 + 8(2)\right] ] Simplify: [ \left[-\frac{64}{3} + 16 + 32\right] - \left[-\frac{8}{3} + 4 + 16\right] ] $\left(\frac{-64 + 48 + 96}{3}\right) - \left(\frac{-8 + 12 + 48}{3}\right)$ $\frac{80}{3} - \frac{52}{3} = \frac{28}{3}$

   For $[4, 5]$: $\int_{4}^{5} (t^2 - 2t - 8) \, dt = \left[\frac{t^3}{3} - t^2 - 8t\right]_{4}^{5}$ Evaluate: [ \left[\frac{5^3}{3} - 5^2 - 8(5)\right] - \left[\frac{4^3}{3} - 4^2 - 8(4)\right] ] Simplify: [ \left[\frac{125}{3} - 25 - 40\right] - \left[\frac{64}{3} - 16 - 32\right] ] $\left(\frac{125 - 75 - 120}{3}\right) - \left(\frac{64 - 48 - 96}{3}\right)$ $\frac{-70}{3} - \frac{-80}{3} = \frac{10}{3}$

5. Add the results: $\text{Total Distance} = \frac{28}{3} + \frac{10}{3} = \frac{38}{3} \, \text{meters}.$

Final Answer:

The distance traveled is: $\boxed{\frac{38}{3} \, \text{meters}}.$

Would you like a further breakdown of the integration steps? Here are some questions to consider:

1. How does the displacement differ conceptually and mathematically from the distance traveled?
2. Can we use symmetry properties of the velocity function to simplify absolute value integrals?
3. What does a velocity function's sign change signify in terms of motion?
4. How does integrating a velocity function relate to real-world applications like speedometers?
5. What tools can automate these computations, especially absolute value integrals?

Tip: Always sketch the velocity function on the interval to visualize the areas contributing to distance. This helps avoid sign errors.