To find the displacement of the particle over the given time interval \(2 \leq t \leq 8\) with the velocity function \(v(t) = t^2 - 2t - 8\), we can use the formula for displacement, which is the integral of the velocity function over the specified interval:
\[ \text{Displacement} = \int_{a}^{b} v(t) , dt \]
where \(a = 2\) and \(b = 8\).
First, let's set up the integral:
\[ \text{Displacement} = \int_{2}^{8} (t^2 - 2t - 8) , dt \]
Next, we can compute the integral:
- Find the antiderivative of \(v(t) = t^2 - 2t - 8\):
\[ \int (t^2 - 2t - 8) , dt = \frac{t^3}{3} - t^2 - 8t + C \]
- Now, we will evaluate the definite integral from 2 to 8:
\[ \int_{2}^{8} (t^2 - 2t - 8) , dt = \left[ \frac{t^3}{3} - t^2 - 8t \right]_{2}^{8} \]
- Calculate the upper limit (when \(t = 8\)):
\[ \frac{8^3}{3} - 8^2 - 8 \cdot 8 = \frac{512}{3} - 64 - 64 \] \[ = \frac{512}{3} - \frac{192}{3} - \frac{192}{3} = \frac{512 - 384}{3} = \frac{128}{3} \]
- Now calculate the lower limit (when \(t = 2\)):
\[ \frac{2^3}{3} - 2^2 - 8 \cdot 2 = \frac{8}{3} - 4 - 16 \] \[ = \frac{8}{3} - \frac{12}{3} - \frac{48}{3} = \frac{8 - 12 - 48}{3} = \frac{8 - 60}{3} = \frac{-52}{3} \]
- Now, subtract the lower limit from the upper limit:
\[ \text{Displacement} = \left( \frac{128}{3} - \left(-\frac{52}{3}\right) \right) = \frac{128}{3} + \frac{52}{3} = \frac{180}{3} = 60 \]
Thus, the displacement of the particle is:
\[ \text{Displacement} = 60 , \text{meters} \]
Your assertion is correct, the displacement is indeed \(60\) meters.
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