To find the distance traveled by the particle from \( t = 0 \) to \( t = 3 \) seconds, we need to integrate the velocity function over that time interval. The given velocity function is:
\[
V(t) = 3t^2 - 18t + 24 \, \text{m/s}
\]
### Step 1: Determine the time intervals where the velocity is positive or negative.
First, we can find the roots of the velocity equation to identify when the particle changes direction:
\[
3t^2 - 18t + 24 = 0
\]
Dividing the entire equation by 3 gives:
\[
t^2 - 6t + 8 = 0
\]
Now, we can factor this quadratic:
\[
(t - 2)(t - 4) = 0
\]
Thus, the roots are:
\[
t = 2 \quad \text{and} \quad t = 4
\]
Since we are only interested in the interval from \( t = 0 \) to \( t = 3 \), we will analyze the sign of \( V(t) \) in the intervals \( [0, 2] \) and \( [2, 3] \).
- For \( t = 1 \) (in the interval \( [0, 2] \)):
\[
V(1) = 3(1)^2 - 18(1) + 24 = 3 - 18 + 24 = 9 \, \text{(positive)}
\]
- For \( t = 3 \) (in the interval \( [2, 3] \)):
\[
V(3) = 3(3)^2 - 18(3) + 24 = 27 - 54 + 24 = -3 \, \text{(negative)}
\]
### Step 2: Set up the integral for distance.
Since the particle changes direction at \( t = 2 \), we need to split the integral into two parts:
\[
\text{Distance} = \int_{0}^{2} V(t) \, dt + \int_{2}^{3} |V(t)| \, dt
\]
### Step 3: Calculate the first integral from \( 0 \) to \( 2 \).
\[
\int_{0}^{2} (3t^2 - 18t + 24) \, dt
\]
Calculating the integral:
\[
= \left[ t^3 - 9t^2 + 24t \right]_{0}^{2}
\]
Calculating at the limits:
\[
= \left[ (2)^3 - 9(2)^2 + 24(2) \right] - \left[ (0)^3 - 9(0)^2 + 24(0) \right]
\]
\[
= \left[ 8 - 36 + 48 \right] - 0
\]
\[
= 20 \, \text{meters}
\]
### Step 4: Calculate the second integral from \( 2 \) to \( 3 \).
Since \( V(t) \) is negative in this interval, we take the absolute value:
\[
\int_{2}^{3} |V(t)| \, dt = -\int_{2}^{3} (3t^2 - 18t + 24) \, dt
\]
Calculating the integral:
\[
= -\left[ t^3 - 9t^2 + 24t \right]_{2}^{3}
\]
Calculating at the limits:
\[
= -\left[ (3)^3 - 9(3)^2 + 24(3) - ((2)^3 - 9(2)^2 + 24(2)) \right]
\]
\[
= -\left[ 27 - 81 + 72 - (8 - 36 + 48) \right]
\]
\[
= -\left[ 27 - 81 + 72 - 20 \right]
\]
\[
= -\left[ -2 \right] = 2 \, \text{meters}
\]
### Step 5: Combine the distances.
Now, we add the distances from both intervals:
\[
\text{Total Distance} = 20 + 2 = 22 \, \text{meters}
\]
### Final Answer:
The distance traveled by the particle from \( t = 0 \) to \( t = 3 \) seconds is \( 22 \, \text{meters} \).
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