Please try again. Because the particle changes direction during the time interval, the displacement will not be equal to the distance traveled. The displacement is simply the integral of the velocity function. The function for total distance traveled, however, treats any motion as positive, so it is the integral of the velocity function's absolute value.


You're correct! When the particle changes direction, the displacement and the distance traveled are indeed different. Let's carefully go through the calculations again.

### Part (a): Displacement

The displacement is the net change in position, which we find by integrating the velocity function over the given time interval \([2, 7]\).

The velocity function is:
\[
v(t) = t^2 - 2t - 8
\]

The displacement \(D\) is given by:
\[
D = \int_{2}^{7} v(t) \, dt
\]

We'll compute this integral:

\[
D = \int_{2}^{7} (t^2 - 2t - 8) \, dt
\]

Let's find the antiderivative:

\[
\text{Antiderivative of } t^2 = \frac{t^3}{3}
\]
\[
\text{Antiderivative of } -2t = -t^2
\]
\[
\text{Antiderivative of } -8 = -8t
\]

So, the antiderivative of \(v(t)\) is:
\[
\frac{t^3}{3} - t^2 - 8t
\]

Now, evaluate this from \(t = 2\) to \(t = 7\):

\[
D = \left[\frac{7^3}{3} - 7^2 - 8(7)\right] - \left[\frac{2^3}{3} - 2^2 - 8(2)\right]
\]

Substituting the values:

\[
D = \left[\frac{343}{3} - 49 - 56\right] - \left[\frac{8}{3} - 4 - 16\right]
\]
\[
D = \left[\frac{343 - 315}{3}\right] - \left[\frac{8 - 60}{3}\right]
\]
\[
D = \left[\frac{28}{3}\right] - \left[\frac{-52}{3}\right]
\]
\[
D = \frac{28 + 52}{3} = \frac{80}{3} \approx 26.67 \text{ meters}
\]

### Part (b): Distance Traveled

The distance traveled takes into account the direction change by integrating the absolute value of the velocity function.

1. **Find when the velocity is zero (where the particle changes direction):**

\[
v(t) = t^2 - 2t - 8 = 0
\]

Factor the quadratic equation:
\[
(t - 4)(t + 2) = 0
\]

So, \(t = 4\) and \(t = -2\). Since \(t = -2\) is outside the interval, only \(t = 4\) is relevant.

2. **Calculate the distance traveled:**

The total distance traveled is the sum of the absolute value of the integrals over the intervals \([2, 4]\) and \([4, 7]\).

\[
\text{Distance} = \left|\int_{2}^{4} v(t) \, dt\right| + \left|\int_{4}^{7} v(t) \, dt\right|
\]

Calculate these integrals:

- **For \([2, 4]\):**
  \[
  \int_{2}^{4} (t^2 - 2t - 8) \, dt
  \]
  Substituting the antiderivatives and evaluating from 2 to 4:
  \[
  = \left[\frac{64}{3} - 16 - 32\right] - \left[\frac{8}{3} - 4 - 16\right]
  \]
  \[
  = \frac{64 - 144}{3} = \frac{-80}{3} \text{ meters}
  \]

- **For \([4, 7]\):**
  \[
  \int_{4}^{7} (t^2 - 2t - 8) \, dt
  \]
  Evaluate this similarly:
  \[
  = 36 \text{ meters}
  \]

Since we're calculating the total distance, we take the absolute values:

\[
\text{Distance} = \left| -\frac{80}{3} \right| + 36 = \frac{80}{3} + 36 \approx 26.67 + 36 = 62.67 \text{ meters}
\]

### Summary:

- **Displacement**: \(\approx 26.67\) meters
- **Distance Traveled**: \(\approx 62.67\) meters

Would you like more details on any part of these calculations or have any questions?

Here are some related questions for further practice:

1. How do you determine when a particle changes direction using a velocity function?
2. What is the physical meaning of integrating the velocity function?
3. How would you find the maximum displacement of a particle over a given interval?
4. What would the distance be if the velocity function was always positive?
5. How does the concept of absolute value affect the distance calculation?

**Tip**: When calculating total distance traveled, always check for sign changes in the velocity function within the interval.

Explore the concepts of displacement and distance traveled using velocity functions. This detailed explanation covers calculus concepts such as integration and absolute value, applied to real-world motion scenarios. Learn how to calculate displacement and total distance traveled using the velocity function's integral. Suitable for advanced high school students and calculus enthusiasts.

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