Certainly! Let's consider the scenario where an officer is accelerating at a constant rate. We are provided with the officer's acceleration and need to determine the distance she travels during a specific period.
### Step-by-Step Solution:
1. Given Data:
- Acceleration of the officer, [tex]\(a = 4.8 \text{ m/s}^2\)[/tex]
- Initial velocity of the officer, [tex]\(u = 0 \text{ m/s}\)[/tex] (assuming the officer started from rest)
- Time, [tex]\(t = 10 \text{ seconds}\)[/tex] (an arbitrary time frame to calculate the distance traveled)
2. Formula for Distance with Constant Acceleration:
We will use the kinematic equation for distance traveled under constant acceleration:
[tex]\[
s = ut + \frac{1}{2}at^2
\][/tex]
Where:
- [tex]\(s\)[/tex] is the distance traveled
- [tex]\(u\)[/tex] is the initial velocity
- [tex]\(a\)[/tex] is the acceleration
- [tex]\(t\)[/tex] is the time
3. Substitute the Given Values:
Plugging in the values from the given data into the formula:
[tex]\[
u = 0 \text{ m/s}
\][/tex]
[tex]\[
a = 4.8 \text{ m/s}^2
\][/tex]
[tex]\[
t = 10 \text{ seconds}
\][/tex]
The equation becomes:
[tex]\[
s = (0 \text{ m/s}) (10 \text{ s}) + \frac{1}{2} (4.8 \text{ m/s}^2) (10 \text{ s})^2
\][/tex]
4. Calculate the Distance:
First, calculate the second term:
[tex]\[
\frac{1}{2} \times 4.8 \text{ m/s}^2 \times (10 \text{ s})^2
\][/tex]
[tex]\[
\frac{1}{2} \times 4.8 \times 100
\][/tex]
[tex]\[
\frac{1}{2} \times 480
\][/tex]
[tex]\[
240 \text{ meters}
\][/tex]
Thus, the distance the officer travels before catching up with the car is:
[tex]\[
\boxed{240 \text{ meters}}
\][/tex]
