Question

The velocity function (in meters per second) is given for a particle moving along a line.
v(t) = 3t − 7, 0 ≤ t ≤ 3
(a)
Find the displacement (in meters).
m
(b)
Find the total distance traveled (in meters) by the particle during the given time interval.
m

Answer 1

To find the displacement and total distance traveled by the particle with the given velocity function \( v(t) = 3t - 7 \) over the interval \( 0 \leq t \leq 3 \), we proceed with the following steps:

### (a) Find the Displacement

Displacement can be found by integrating the velocity function over the given interval:

\[
\text{Displacement} = \int_{0}^{3} v(t) \, dt = \int_{0}^{3} (3t - 7) \, dt
\]

Now, we calculate the integral:

1. Find the antiderivative:
\[
\int (3t - 7) \, dt = \frac{3t^2}{2} - 7t
\]

2. Evaluate the definite integral from \( t = 0 \) to \( t = 3 \):
\[
\left[ \frac{3(3)^2}{2} - 7(3) \right] - \left[ \frac{3(0)^2}{2} - 7(0) \right]
\]

Simplifying this:
\[
\left[ \frac{3 \cdot 9}{2} - 21 \right] - [0]
\]
\[
= \left[ \frac{27}{2} - 21 \right]
\]
\[
= \frac{27}{2} - \frac{42}{2} = \frac{-15}{2}
\]

Thus, the displacement is:
\[
\text{Displacement} = -\frac{15}{2} \text{ meters} = -7.5 \text{ meters}
\]

### (b) Find the Total Distance Traveled

To find the total distance traveled, we need to determine when the velocity changes sign, indicating a change in direction. We set the velocity function to zero:

\[
3t - 7 = 0 \implies 3t = 7 \implies t = \frac{7}{3} \approx 2.33
\]

Now, we split the integral at \( t = \frac{7}{3} \) since the velocity is negative when \( t < \frac{7}{3} \) and positive when \( t > \frac{7}{3} \).

1. **From \( 0 \) to \( \frac{7}{3} \)** (where \( v(t) < 0 \)):
\[
\text{Distance}_1 = -\int_{0}^{\frac{7}{3}} v(t) \, dt = -\int_{0}^{\frac{7}{3}} (3t - 7) \, dt
\]
The antiderivative is the same:
\[
= -\left[ \frac{3t^2}{2} - 7t \right]_{0}^{\frac{7}{3}}
\]
Calculating this gives:
\[
= -\left[ \frac{3(\frac{7}{3})^2}{2} - 7\left(\frac{7}{3}\right) \right]
\]
\[
= -\left[ \frac{3 \cdot \frac{49}{9}}{2} - \frac{49}{3} \right]
\]
\[
= -\left[ \frac{147}{18} - \frac{294}{18} \right] = -\left[ -\frac{147}{18} \right] = \frac{147}{18} = \frac{49}{6}
\]

2. **From \( \frac{7}{3} \) to \( 3 \)** (where \( v(t) > 0 \)):
\[
\text{Distance}_2 = \int_{\frac{7}{3}}^{3} v(t) \, dt = \int_{\frac{7}{3}}^{3} (3t - 7) \, dt
\]
This is calculated as follows:
\[
= \left[ \frac{3t^2}{2} - 7t \right]_{\frac{7}{3}}^{3}
\]
Evaluating this gives:
\[
= \left[ \frac{3(3)^2}{2} - 7(3) \right] - \left[ \frac{3(\frac{7}{3})^2}{2} - 7\left(\frac{7}{3}\right) \right]
\]
The first term is \( \frac{-15}{2} \) (calculated above) and the second term is \( -\frac{49}{6} \).

The total distance traveled is then:

\[
\text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 = \frac{49}{6} + \frac{49}{6} = \frac{98}{6} = \frac{49}{3} \text{ meters} \approx 16.33 \text{ meters}
\]

### Summary of Results
(a) Displacement: \(-7.5\) meters

(b) Total distance traveled: \(\frac{49}{3} \approx 16.33\) meters