To find the displacement and total distance traveled over the time interval $[0, 10]$, we need to analyze the velocity function $v(t) = 3t^2 - 24t + 45$.

Step 1: Displacement

Displacement is the integral of the velocity function over the interval $[0, 10]$.

$\text{Displacement} = \int_{0}^{10} v(t) \, dt = \int_{0}^{10} (3t^2 - 24t + 45) \, dt$

$\int (3t^2 - 24t + 45) \, dt = t^3 - 12t^2 + 45t + C$

Evaluating this from 0 to 10:

$\text{Displacement} = \left[ t^3 - 12t^2 + 45t \right]_0^{10}$

At $t = 10$: $(10)^3 - 12(10)^2 + 45(10) = 1000 - 1200 + 450 = 250 \, \text{m}$

At $t = 0$: $(0)^3 - 12(0)^2 + 45(0) = 0$

Thus, the displacement is: $\text{Displacement} = 250 - 0 = 250 \, \text{m}.$

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Step 2: Total Distance Traveled

To find the total distance traveled, we need to account for changes in the direction of motion (when $v(t)$ changes sign). This requires finding the roots of $v(t)$.

$v(t) = 3t^2 - 24t + 45 = 0$

Divide by 3: $t^2 - 8t + 15 = 0$

Factorize: $(t - 5)(t - 3) = 0$

The roots are $t = 3$ and $t = 5$. Thus, the velocity changes sign at $t = 3$ and $t = 5$. We split the interval $[0, 10]$ into three subintervals: $[0, 3]$, $[3, 5]$, and $[5, 10]$.

1. Interval $[0, 3]$: $v(t) > 0$, so the distance is the same as the displacement: $\int_{0}^{3} v(t) \, dt = \left[ t^3 - 12t^2 + 45t \right]_0^3$

   At $t = 3$: $(3)^3 - 12(3)^2 + 45(3) = 27 - 108 + 135 = 54 \, \text{m}.$

   At $t = 0$: $(0)^3 - 12(0)^2 + 45(0) = 0.$

   Distance for $[0, 3]$: $54 - 0 = 54 \, \text{m}.$

2. Interval $[3, 5]$: $v(t) < 0$, so we take the absolute value of the displacement: $\int_{3}^{5} |v(t)| \, dt = -\int_{3}^{5} v(t) \, dt$

   $\int_{3}^{5} v(t) \, dt = \left[ t^3 - 12t^2 + 45t \right]_3^5$

   At $t = 5$: $(5)^3 - 12(5)^2 + 45(5) = 125 - 300 + 225 = 50 \, \text{m}.$

   At $t = 3$: $(3)^3 - 12(3)^2 + 45(3) = 54 \, \text{m}.$

   Distance for $[3, 5]$: $|50 - 54| = 4 \, \text{m}.$

3. Interval $[5, 10]$: $v(t) > 0$, so the distance is the same as the displacement: $\int_{5}^{10} v(t) \, dt = \left[ t^3 - 12t^2 + 45t \right]_5^{10}$

   At $t = 10$: $(10)^3 - 12(10)^2 + 45(10) = 250 \, \text{m}.$

   At $t = 5$: $(5)^3 - 12(5)^2 + 45(5) = 50 \, \text{m}.$

   Distance for $[5, 10]$: $250 - 50 = 200 \, \text{m}.$

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Total Distance:

$\text{Total Distance} = 54 + 4 + 200 = 258 \, \text{m}.$

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Final Answers:

• Displacement: $250 \, \text{m}$
• Total Distance Traveled: $258 \, \text{m}$

Let me know if you need detailed graphs or further explanations!

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Relative Questions:

1. What is the significance of finding displacement in motion analysis?
2. How do you interpret a velocity function geometrically on a graph?
3. Why is the total distance different from displacement for motions with changing direction?
4. How can we use definite integrals to compute physical quantities like displacement or distance?
5. What additional parameters could we calculate from $v(t)$, such as acceleration or speed?

Tip: Always verify the points of velocity sign change when computing total distance to avoid errors!