Question

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = t2 − 2t − 15, 2 ≤ t ≤ 8
(a)
Find the displacement (in meters).
m
(b)
Find the distance traveled (in meters) by the particle during the given time interval.
m

Answer 1

To solve the problem, we need to work with the velocity function \( v(t) = t^2 - 2t - 15 \). We’ll perform two tasks: find the displacement and then find the distance traveled by the particle from \( t = 2 \) to \( t = 8 \).

### (a) Finding the Displacement

The displacement over the time interval can be found by integrating the velocity function:

\[
\text{Displacement} = \int_{2}^{8} v(t) \, dt = \int_{2}^{8} (t^2 - 2t - 15) \, dt
\]

First, we compute the integral:

\[
\int (t^2 - 2t - 15) \, dt = \frac{t^3}{3} - t^2 - 15t + C
\]

Now, we evaluate this from \( t = 2 \) to \( t = 8 \):

\[
\left[ \frac{t^3}{3} - t^2 - 15t \right]_{2}^{8}
\]

Calculating at \( t = 8 \):

\[
\frac{8^3}{3} - 8^2 - 15(8) = \frac{512}{3} - 64 - 120
\]
\[
= \frac{512}{3} - \frac{192}{3} - \frac{360}{3} = \frac{512 - 192 - 360}{3} = \frac{-40}{3}
\]

Now calculating at \( t = 2 \):

\[
\frac{2^3}{3} - 2^2 - 15(2) = \frac{8}{3} - 4 - 30 = \frac{8}{3} - \frac{12}{3} - \frac{90}{3} = \frac{8 - 12 - 90}{3} = \frac{-94}{3}
\]

Now we find the displacement:

\[
\text{Displacement} = \left( \frac{-40}{3} \right) - \left( \frac{-94}{3} \right) = \frac{-40 + 94}{3} = \frac{54}{3} = 18 \text{ meters}
\]

Thus, the displacement is:

\[
\boxed{18} \text{ meters}
\]

### (b) Finding the Distance Traveled

To find the distance traveled, we need to take into account when the velocity changes sign within the interval \( [2, 8] \). We first find the roots of the velocity function \( v(t) = 0 \):

\[
t^2 - 2t - 15 = 0
\]

Using the quadratic formula:

\[
t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)} = \frac{2 \pm \sqrt{4 + 60}}{2} = \frac{2 \pm \sqrt{64}}{2} = \frac{2 \pm 8}{2}
\]

Calculating the roots:

\[
t = \frac{10}{2} = 5 \quad \text{and} \quad t = \frac{-6}{2} = -3 \text{ (not in the interval)}
\]

The velocity changes sign at \( t = 5 \). We evaluate the distance by integrating the absolute value of the velocity in the intervals \( [2, 5] \) and \( [5, 8] \):

1. For \( t \in [2, 5] \): \( v(t) \) is negative since \( v(4) = 4 - 8 - 15 < 0 \).

2. For \( t \in [5, 8] \): \( v(t) \) is positive since \( v(6) = 36 - 12 - 15 > 0 \).

Now we compute:

\[
\text{Distance} = \int_{2}^{5} -v(t) \, dt + \int_{5}^{8} v(t) \, dt
\]

Calculating the first integral:

\[
\int_{2}^{5} -v(t) \, dt = \int_{2}^{5} -(t^2 - 2t - 15) \, dt = \int_{2}^{5} (-t^2 + 2t + 15) \, dt
\]

Calculating the integral:

\[
\int (-t^2 + 2t + 15) \, dt = -\frac{t^3}{3} + t^2 + 15t + C
\]

Evaluate from 2 to 5:

\[
\left[-\frac{t^3}{3} + t^2 + 15t \right]_{2}^{5}
\]

Calculating at \( t = 5 \):

\[
-\frac{5^3}{3} + 5^2 + 15(5) = -\frac{125}{3} + 25 + 75 = -\frac{125}{3} + 100 = \frac{300 - 125}{3} = \frac{175}{3}
\]

Calculating at \( t = 2 \):

\[
-\frac{2^3}{3} + 2^2 + 15(2) = -\frac{8}{3} + 4 + 30 = -\frac{8}{3} + 34 = \frac{102 - 8}{3} = \frac{94}{3}
\]

Now evaluate the first integral:

\[
\int_{2}^{5} -v(t) \, dt = \left( \frac{175}{3} - \frac{94}{3} \right) = \frac{81}{3} = 27
\]

Now for the second integral from \( t = 5 \) to \( t = 8 \):

\[
\int_{5}^{8} v(t) \, dt = \int_{5}^{8} (t^2 - 2t - 15) \, dt
\]

We already computed the first part up to \( t = 8 \) earlier. So,

\[
\text{Distance} = \int_{2}^{5} -v(t) \, dt + \int_{5}^{8} v(t) \, dt = 27 + \left( \frac{-40}{3} + \frac{94}{3} \right) = 27 + \frac{54}{3} = 27 + 18 = 45
\]

Thus, the total distance traveled is:

\[
\boxed{45} \text{ meters}
\]