To solve the problem regarding the motion of the particle defined by the position function \( x(t) = \frac{t^3}{3} - 3t^2 + 8t + 4 \) over the interval from \( t = 0 \) to \( t = 5 \) seconds, we need to analyze the particle's motion, particularly focusing on the total distance traveled (S1) and the distance traveled during periods of retardation (S2).
Step 1: Finding the Velocity Function
First, we need to determine the velocity of the particle by differentiating the position function with respect to time:
Velocity, v(t) = x'(t) = \frac{d}{dt}\left(\frac{t^3}{3} - 3t^2 + 8t + 4\right)
Calculating the derivative:
- Derivative of \( \frac{t^3}{3} \) is \( t^2 \)
- Derivative of \( -3t^2 \) is \( -6t \)
- Derivative of \( 8t \) is \( 8 \)
- Derivative of constant \( 4 \) is \( 0 \)
Thus, the velocity function is:
v(t) = t^2 - 6t + 8
Step 2: Analyzing the Velocity Function
Next, we need to find the critical points where the velocity is zero, as this will help us identify when the particle changes direction:
Setting \( v(t) = 0 \):
t^2 - 6t + 8 = 0
Factoring the quadratic gives:
(t - 2)(t - 4) = 0
Thus, the critical points are \( t = 2 \) seconds and \( t = 4 \) seconds.
Step 3: Evaluating the Position at Critical Points
Now, we will evaluate the position function at \( t = 0, 2, 4, \) and \( 5 \) seconds to find the total distance traveled:
- x(0) = 4
- x(2) = \frac{8}{3} - 12 + 16 + 4 = \frac{8}{3} + 8 = \frac{32}{3} \approx 10.67
- x(4) = \frac{64}{3} - 48 + 32 + 4 = \frac{64}{3} - 12 = \frac{8}{3} \approx 2.67
- x(5) = \frac{125}{3} - 75 + 40 + 4 = \frac{125}{3} - 31 = \frac{32}{3} \approx 10.67
Step 4: Calculating Total Distance (S1)
The total distance traveled, S1, is the sum of the absolute differences in position between these points:
- From \( t = 0 \) to \( t = 2 \): \( |x(2) - x(0)| = \left| \frac{32}{3} - 4 \right| = \frac{32}{3} - \frac{12}{3} = \frac{20}{3} \)
- From \( t = 2 \) to \( t = 4 \): \( |x(4) - x(2)| = \left| \frac{8}{3} - \frac{32}{3} \right| = \frac{24}{3} = 8 \)
- From \( t = 4 \) to \( t = 5 \): \( |x(5) - x(4)| = \left| \frac{32}{3} - \frac{8}{3} \right| = \frac{24}{3} = 8 \)
Thus, the total distance \( S1 \) is:
S1 = \frac{20}{3} + 8 + 8 = \frac{20}{3} + \frac{24}{3} + \frac{24}{3} = \frac{68}{3}
Step 5: Calculating Distance During Retardation (S2)
Retardation occurs when the velocity is positive and then becomes negative. This happens between \( t = 2 \) and \( t = 4 \). Thus, S2 is the distance traveled during this interval:
S2 = |x(4) - x(2)| = 8
Step 6: Finding the Value of α
We know from the problem statement that:
\( \frac{S1}{S2} = \frac{3\alpha + 2}{11} \)
Substituting the values we found:
\( \frac{\frac{68}{3}}{8} = \frac{3\alpha + 2}{11} \)
This simplifies to:
\( \frac{68}{24} = \frac{3\alpha + 2}{11} \)
Cross-multiplying gives:
68 * 11 = 24(3\alpha + 2)
Calculating this results in:
748 = 72\alpha + 48
Solving for α:
72\alpha = 700
α = \frac{700}{72} = \frac{175}{18} \approx 9.72
Thus, the value of α is approximately 9.72.
