Let’s solve the problem step by step to find the displacement and total distance traveled.

(a) Displacement

Displacement is calculated using the definite integral of the velocity function $v(t)$ over the given time interval $[0, 4]$.

$\text{Displacement} = \int_{0}^{4} v(t) \, dt = \int_{0}^{4} (3t - 8) \, dt$

1. Compute the integral: $\int (3t - 8) \, dt = \frac{3t^2}{2} - 8t + C$

2. Apply the limits of integration: [ \text{Displacement} = \left[ \frac{3(4)^2}{2} - 8(4) \right] - \left[ \frac{3(0)^2}{2} - 8(0) \right] ] [ \text{Displacement} = \left[ \frac{48}{2} - 32 \right] - \left[ 0 \right] = 24 - 32 = -8 , \text{m}. ]

This confirms your correct answer of $-8 \, \text{m}$ for displacement.

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(b) Total Distance Traveled

Total distance differs from displacement because it considers the absolute value of the velocity $|v(t)|$.

1. Identify when the velocity changes sign, since the particle may reverse direction: $v(t) = 3t - 8 \quad \text{set } v(t) = 0$ $3t - 8 = 0 \implies t = \frac{8}{3} \, \text{seconds}.$

   So, the particle changes direction at $t = \frac{8}{3}$. We’ll split the integral into two parts: $[0, \frac{8}{3}]$ and $[\frac{8}{3}, 4]$.

2. Compute the total distance: $\text{Total Distance} = \int_{0}^{\frac{8}{3}} |v(t)| \, dt + \int_{\frac{8}{3}}^{4} |v(t)| \, dt$

For $0 \leq t \leq \frac{8}{3}$, $v(t) < 0$, so $|v(t)| = -(3t - 8) = -3t + 8$.

For $\frac{8}{3} \leq t \leq 4$, $v(t) > 0$, so $|v(t)| = 3t - 8$.

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Step 1: Compute the first integral $\int_{0}^{\frac{8}{3}} (-3t + 8) \, dt = \left[ -\frac{3t^2}{2} + 8t \right]_{0}^{\frac{8}{3}}$ [ = \left[ -\frac{3\left(\frac{8}{3}\right)^2}{2} + 8\left(\frac{8}{3}\right) \right] - \left[ 0 \right] ] $= \left[ -\frac{3 \cdot \frac{64}{9}}{2} + \frac{64}{3} \right]$ $= \left[ -\frac{192}{18} + \frac{384}{18} \right] = \frac{192}{18} = \frac{32}{3} \, \text{m}.$

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Step 2: Compute the second integral $\int_{\frac{8}{3}}^{4} (3t - 8) \, dt = \left[ \frac{3t^2}{2} - 8t \right]_{\frac{8}{3}}^{4}$ [ = \left[ \frac{3(4)^2}{2} - 8(4) \right] - \left[ \frac{3\left(\frac{8}{3}\right)^2}{2} - 8\left(\frac{8}{3}\right) \right] ] [ = \left[ 24 - 32 \right] - \left[ -\frac{192}{18} + \frac{384}{18} \right] ] $= -8 - \frac{192}{18} = \frac{32}{3} \, \text{m}.$

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Step 3: Total Distance $\text{Total Distance} = \frac{32}{3} + \frac{32}{3} = \frac{64}{3} \approx 21.33 \, \text{m}.$

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Final Answers:

(a) Displacement: $-8 \, \text{m}$
(b) Total Distance Traveled: $\frac{64}{3} \, \text{m} \approx 21.33 \, \text{m}$

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Would you like more details on the calculations or further assistance?
Here are 5 related questions to expand your understanding:

1. How is displacement different from distance traveled in motion problems?
2. What does the velocity sign change signify about the particle's motion?
3. How can you graph velocity to visually interpret displacement and distance?
4. What is the significance of using absolute value in total distance calculations?
5. Can total distance ever equal displacement? If so, under what conditions?

Tip: When solving for total distance, always split integrals at points where the velocity changes sign!