Question

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t2 − 2t − 8, 2 ≤ t ≤ 6
(a)
Find the displacement (in meters).
m
(b)
Find the distance traveled (in meters) by the particle during the given time interval.
m
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Answer 1

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:

1. **Find the antiderivative**:

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

2. **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 \( t = 2 \):

\[
\frac{2^3}{3} - 2^2 - 8 \cdot 2 = \frac{8}{3} - 4 - 16 = \frac{8}{3} - 12 = \frac{8 - 36}{3} = \frac{-28}{3}
\]

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.

1. **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:

\[
t = 4 \quad (\text{since } t = -2 \text{ is not in the interval})
\]

2. **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:

\[
\boxed{\frac{64}{3}} \text{ meters}
\]