A car is speeding at 25 m/s (constant) in a school zone. A police car starts from rest just as the speeder passes and accelerates at a constant rate of 5.0 m/s². How fast is the police car traveling when it catches up with the speeder?

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### Physics Problem: Catching Up with a Speeder

**Problem Statement:**

A car is speeding at 25 m/s (constant) in a school zone. A police car starts from rest just as the speeder passes and accelerates at a constant rate of 5.0 m/s². How fast is the police car traveling when it catches up with the speeder?

**Analysis Steps:**

1. **Identify Given Data:**
   - Speed of the speeder: \( v_s = 25 \, \text{m/s} \) (constant)
   - Initial speed of the police car: \( v_{p0} = 0 \, \text{m/s} \)
   - Acceleration of the police car: \( a_p = 5.0 \, \text{m/s}^2 \)

2. **Determine the time it takes for the police car to catch up:**
   - The distance traveled by the speeder in time \( t \): 
     \[
     d_s = v_s \cdot t
     \]
   - The distance traveled by the police car in time \( t \):
     \[
     d_p = \frac{1}{2} a_p \cdot t^2
     \]
   - Set \( d_s = d_p \) to find \( t \):
     \[
     v_s \cdot t = \frac{1}{2} a_p \cdot t^2
     \]
     \[
     25 \cdot t = \frac{1}{2} \cdot 5.0 \cdot t^2
     \]
     \[
     25t = 2.5t^2
     \]
     \[
     t = \frac{25}{2.5} = 10 \, \text{seconds}
     \]

3. **Calculate the final velocity of the police car:**
   - Use the equation of motion:
     \[
     v_p = v_{p0} + a_p \cdot t
     \]
     \[
     v_p = 0 + 5.0 \cdot 10 = 50 \, \text{m/s}
     \]

**Answer:**

When the police car catches up with the speeder, it is traveling at a speed of \( 50 \, \
Transcribed Image Text:### Physics Problem: Catching Up with a Speeder **Problem Statement:** A car is speeding at 25 m/s (constant) in a school zone. A police car starts from rest just as the speeder passes and accelerates at a constant rate of 5.0 m/s². How fast is the police car traveling when it catches up with the speeder? **Analysis Steps:** 1. **Identify Given Data:** - Speed of the speeder: \( v_s = 25 \, \text{m/s} \) (constant) - Initial speed of the police car: \( v_{p0} = 0 \, \text{m/s} \) - Acceleration of the police car: \( a_p = 5.0 \, \text{m/s}^2 \) 2. **Determine the time it takes for the police car to catch up:** - The distance traveled by the speeder in time \( t \): \[ d_s = v_s \cdot t \] - The distance traveled by the police car in time \( t \): \[ d_p = \frac{1}{2} a_p \cdot t^2 \] - Set \( d_s = d_p \) to find \( t \): \[ v_s \cdot t = \frac{1}{2} a_p \cdot t^2 \] \[ 25 \cdot t = \frac{1}{2} \cdot 5.0 \cdot t^2 \] \[ 25t = 2.5t^2 \] \[ t = \frac{25}{2.5} = 10 \, \text{seconds} \] 3. **Calculate the final velocity of the police car:** - Use the equation of motion: \[ v_p = v_{p0} + a_p \cdot t \] \[ v_p = 0 + 5.0 \cdot 10 = 50 \, \text{m/s} \] **Answer:** When the police car catches up with the speeder, it is traveling at a speed of \( 50 \, \
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