To find the displacement and distance traveled by the particle, we will analyze the velocity function \( v(t) = t^2 - 2t - 8 \) over the interval \( 2 \leq t \leq 6 \).
(a) Finding Displacement
To find the displacement, we need to calculate the definite integral of the velocity function over the given interval:
\[ \text{Displacement} = \int_{2}^{6} v(t) , dt = \int_{2}^{6} (t^2 - 2t - 8) , dt \]
We will compute the integral:
- Find the antiderivative:
\[ \int (t^2 - 2t - 8) , dt = \frac{t^3}{3} - t^2 - 8t + C \]
- Evaluate the definite integral from \( t = 2 \) to \( t = 6 \):
\[ \int_{2}^{6} (t^2 - 2t - 8) , dt = \left[ \frac{t^3}{3} - t^2 - 8t \right]_{2}^{6} \]
Calculating at \( t = 6 \):
\[ \frac{6^3}{3} - 6^2 - 8 \cdot 6 = \frac{216}{3} - 36 - 48 = 72 - 36 - 48 = -12 \]
Calculating at \( t = 2 \):
\[ \frac{2^3}{3} - 2^2 - 8 \cdot 2 = \frac{8}{3} - 4 - 16 = \frac{8}{3} - 12 = \frac{8 - 36}{3} = \frac{-28}{3} \]
- Subtract the two values:
\[ \text{Displacement} = -12 - \left( \frac{-28}{3} \right) = -12 + \frac{28}{3} = -\frac{36}{3} + \frac{28}{3} = -\frac{8}{3} \]
Thus, the displacement is:
\[ \boxed{-\frac{8}{3}} \text{ meters} \]
(b) Finding Distance Traveled
To find the distance traveled, we need to check where the velocity changes sign, as this affects how distance is calculated.
- Find roots of \( v(t) = 0 \):
Solve the quadratic equation:
\[ t^2 - 2t - 8 = 0 \]
Using the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]
This gives us:
\[ t = 4 \quad (\text{since } t = -2 \text{ is not in the interval}) \]
- Evaluate the distance traveled by integrating the absolute value of the velocity between the intervals where \( v(t) \) changes sign:
The velocity \( v(t) \) is negative on \( [2, 4] \) and positive on \( [4, 6] \).
Now calculate:
\[ \text{Distance} = \int_{2}^{4} -v(t) , dt + \int_{4}^{6} v(t) , dt \]
Calculating the first integral:
\[ \int_{2}^{4} -(t^2 - 2t - 8) , dt = \int_{2}^{4} (-t^2 + 2t + 8) , dt \]
Antiderivative:
\[ \int (-t^2 + 2t + 8) , dt = -\frac{t^3}{3} + t^2 + 8t + C \]
Evaluating from \( 2 \) to \( 4 \):
At \( t = 4 \):
\[ -\frac{4^3}{3} + 4^2 + 8 \cdot 4 = -\frac{64}{3} + 16 + 32 = -\frac{64}{3} + 48 = \frac{144 - 64}{3} = \frac{80}{3} \]
At \( t = 2 \):
\[ -\frac{2^3}{3} + 2^2 + 8 \cdot 2 = -\frac{8}{3} + 4 + 16 = -\frac{8}{3} + 20 = \frac{60 - 8}{3} = \frac{52}{3} \]
Thus,
\[ \int_{2}^{4} -v(t) , dt = \frac{80}{3} - \frac{52}{3} = \frac{28}{3} \]
Now, for the second integral:
\[ \int_{4}^{6} v(t) , dt \]
We already computed this and found:
\[ \int_{4}^{6} v(t) , dt = -12 \]
Now, combine the distances:
\[ \text{Distance} = \frac{28}{3} + 12 = \frac{28}{3} + \frac{36}{3} = \frac{64}{3} \]
Thus, the distance traveled is:
\[ \boxed{\frac{64}{3}} \text{ meters} \]
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