The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t2 − 2t − 8,    2 ≤ t ≤ 7
(a)
Find the displacement (in meters).
 
Incorrect: Your answer is incorrect.
 m
(b)
Find the distance traveled (in meters) by the particle during the given time interval.
 
Incorrect: Your answer is incorrect.
 m

To solve the problem, we'll go through both parts step by step. Let's begin with the displacement and then find the total distance traveled by the particle.

### Part (a): Displacement

Displacement is the net change in position of the particle over the time interval. To find the displacement, we need to integrate 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:
\[
D = \int_{2}^{7} v(t) \, dt
\]

Let's compute this integral:

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

To integrate, we find the antiderivative of the function:

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

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

Now, evaluate it from 2 to 7:

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

Let's calculate this.

\[
D = \left[\frac{343}{3} - 49 - 56\right] - \left[\frac{8}{3} - 4 - 16\right]
\]

\[
D = \left[\frac{343}{3} - 105\right] - \left[\frac{8}{3} - 20\right]
\]

\[
D = \frac{343 - 315}{3} - \frac{8 - 60}{3}
\]

\[
D = \frac{28 + 52}{3}
\]

\[
D = 26.67 + 17.33
\]

\[
D = 44 \text{ meters}
\]

### Part (b): Distance Traveled

The distance traveled considers the absolute value of velocity, meaning we account for any change in direction of the particle.

1. First, find when \(v(t) = 0\) within the interval \([2, 7]\):
\[
t^2 - 2t - 8 = 0
\]

Solve this quadratic equation:
\[
t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}
\]
\[
t = 4 \quad \text{or} \quad t = -2 \quad (\text{Not in the interval, so ignore it.})
\]

So, \(v(t)\) changes sign at \(t = 4\).

2. The distance traveled is then the integral of the absolute value of velocity over the intervals \([2, 4]\) and \([4, 7]\):

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

We'll calculate these integrals separately.

For \([2, 4]\):
\[
\int_{2}^{4} (t^2 - 2t - 8) \, dt
\]
Evaluate:
\[
= \left[\frac{4^3}{3} - 4^2 - 8(4)\right] - \left[\frac{2^3}{3} - 2^2 - 8(2)\right]
\]
\[
= \left[ \frac{64}{3} - 16 - 32 \right] - \left[ \frac{8}{3} - 4 - 16 \right]
\]
\[
= \left[ \frac{64 - 144}{3} \right] - \left[\frac{8 - 60}{3}\right]
\]
\[
= 8 \text{ meters}
\]

Now, for \([4, 7]\):

\[
\int_{4}^{7} (t^2 - 2t - 8) \, dt
\]

Evaluate similarly:
\[
D_{4 \rightarrow 7} = 36 \text{ meters}
\]

So the total distance:
\[
\text{Total distance} = 36 \text{ meters}
\]

## Summary

- **Displacement**:  \(44\) meters
- **Distance Traveled**: \(44\) meters

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

Explore how to calculate displacement and distance traveled of a particle using its velocity function. Solve for displacement over a specified time interval and determine total distance traveled, considering direction changes. Detailed step-by-step solution provided.

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