Question

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t2 − 2t − 8, 2 ≤ t ≤ 8
(a)
Find the displacement (in meters).
60
Correct: Your answer is correct.
m

Answer 1

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:

1. Find the antiderivative of \(v(t) = t^2 - 2t - 8\):

\[
\int (t^2 - 2t - 8) \, dt = \frac{t^3}{3} - t^2 - 8t + C
\]

2. Now, we will evaluate the definite integral from 2 to 8:

\[
\int_{2}^{8} (t^2 - 2t - 8) \, dt = \left[ \frac{t^3}{3} - t^2 - 8t \right]_{2}^{8}
\]

3. Calculate the upper limit (when \(t = 8\)):

\[
\frac{8^3}{3} - 8^2 - 8 \cdot 8 = \frac{512}{3} - 64 - 64
\]
\[
= \frac{512}{3} - \frac{192}{3} - \frac{192}{3} = \frac{512 - 384}{3} = \frac{128}{3}
\]

4. Now calculate the lower limit (when \(t = 2\)):

\[
\frac{2^3}{3} - 2^2 - 8 \cdot 2 = \frac{8}{3} - 4 - 16
\]
\[
= \frac{8}{3} - \frac{12}{3} - \frac{48}{3} = \frac{8 - 12 - 48}{3} = \frac{8 - 60}{3} = \frac{-52}{3}
\]

5. 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.