To solve this, we need to use integrals to find the displacement and total distance traveled.

---

(a) Finding the Displacement

Displacement is calculated by integrating the velocity function $v(t)$ over the given time interval:

$\text{Displacement} = \int_{0}^{3} v(t) \, dt = \int_{0}^{3} (3t - 7) \, dt$

1. Integrate $v(t) = 3t - 7$:

$\int (3t - 7) \, dt = \frac{3t^2}{2} - 7t + C$

2. Evaluate the definite integral from $t = 0$ to $t = 3$:

[ \left[ \frac{3(3)^2}{2} - 7(3) \right] - \left[ \frac{3(0)^2}{2} - 7(0) \right] ]

[ = \left[ \frac{27}{2} - 21 \right] - \left[ 0 \right] ]

$= \frac{27}{2} - \frac{42}{2} = -\frac{15}{2} = -7.5 \, \text{meters}.$

Displacement = -7.5 meters.

---

(b) Finding the Total Distance Traveled

The total distance traveled is the integral of the absolute value of the velocity function:

$\text{Total Distance} = \int_{0}^{3} |v(t)| \, dt.$

First, identify when $v(t)$ changes sign by solving $v(t) = 0$:

$3t - 7 = 0 \implies t = \frac{7}{3}.$

So, split the integral at $t = \frac{7}{3}$ because the velocity changes from negative to positive:

$\text{Total Distance} = \int_{0}^{\frac{7}{3}} |v(t)| \, dt + \int_{\frac{7}{3}}^{3} |v(t)| \, dt.$

1. From $t = 0$ to $t = \frac{7}{3}$: $v(t) < 0$, so $|v(t)| = -(v(t)) = -(3t - 7) = -3t + 7$.

$\int_{0}^{\frac{7}{3}} (-3t + 7) \, dt = \left[ -\frac{3t^2}{2} + 7t \right]_{0}^{\frac{7}{3}}.$

[ = \left[ -\frac{3(\frac{7}{3})^2}{2} + 7(\frac{7}{3}) \right] - \left[ 0 \right] ]

$= \left[ -\frac{3(\frac{49}{9})}{2} + \frac{49}{3} \right]$

$= \left[ -\frac{49}{6} + \frac{98}{6} \right] = \frac{49}{6}.$

2. From $t = \frac{7}{3}$ to $t = 3$: $v(t) > 0$, so $|v(t)| = v(t) = 3t - 7$.

$\int_{\frac{7}{3}}^{3} (3t - 7) \, dt = \left[ \frac{3t^2}{2} - 7t \right]_{\frac{7}{3}}^{3}.$

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

[ = \left[ \frac{27}{2} - 21 \right] - \left[ \frac{49}{6} - \frac{49}{3} \right]. ]

Simplify each term:

$\frac{27}{2} - 21 = -\frac{15}{2}, \quad \frac{49}{6} - \frac{98}{6} = -\frac{49}{6}.$

$-\frac{15}{2} - \left( -\frac{49}{6} \right) = -\frac{45}{6} + \frac{49}{6} = \frac{4}{6} = \frac{2}{3}.$

Total Distance Traveled = $\frac{49}{6} + \frac{2}{3} = \frac{51}{6} = 8.5 \, \text{meters}$.

---

Final Answers:

(a) Displacement = $-7.5 \, \text{meters}$
(b) Total Distance Traveled = $8.5 \, \text{meters}$

Would you like further explanation or a breakdown of any step?

---

Related Questions:

1. How does the velocity function affect displacement vs. distance traveled?
2. What does a negative displacement indicate about motion?
3. Can the total distance ever be less than displacement?
4. How do you find when the particle changes direction using velocity?
5. Why is the integral of the absolute value used for total distance?

Tip: When dealing with motion problems, always check for sign changes in velocity to separate the motion into distinct intervals.