An unmarked police car traveling a constant 95.0 km/h
 is passed by a speeder traveling 115 km/h
 .
Part A
Precisely 2.00 s
 after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s2
 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

To solve this problem, we need to determine when the police car overtakes the speeder. Here's a step-by-step breakdown:

### Step 1: Convert speeds from km/h to m/s
We need to convert both the speeds of the speeder and the police car into meters per second (m/s).

- Speeder's speed: 
\[
115 \, \text{km/h} = \frac{115 \times 1000}{3600} \approx 31.94 \, \text{m/s}
\]

- Police car's speed: 
\[
95.0 \, \text{km/h} = \frac{95.0 \times 1000}{3600} \approx 26.39 \, \text{m/s}
\]

### Step 2: Define variables and set up equations
- The speeder moves at a **constant speed** of 31.94 m/s.
- The police car initially travels at 26.39 m/s but begins to accelerate 2.00 seconds after the speeder passes.

Let:
- \( t \) be the time it takes the police car to overtake the speeder **after the speeder passes**.
- The police car begins to accelerate after \( 2.00 \, \text{s} \) at a rate of \( a = 2.40 \, \text{m/s}^2 \).

#### Speeder's Position
The speeder's position at any time \( t \) (starting from \( t = 0 \)) is given by:
\[
x_{\text{speeder}} = v_{\text{speeder}} \cdot t = 31.94 \cdot t
\]

#### Police Car's Position
For the first 2.00 seconds, the police car moves at a constant speed:
\[
x_{\text{police}} = v_{\text{police}} \cdot t = 26.39 \cdot 2.00 = 52.78 \, \text{m}
\]

After the 2.00 seconds, the police car begins to accelerate. So, for \( t > 2.00 \, \text{s} \), the position of the police car is:
\[
x_{\text{police}} = 52.78 + v_{\text{police}} \cdot (t - 2) + \frac{1}{2} a (t - 2)^2
\]
\[
x_{\text{police}} = 52.78 + 26.39 \cdot (t - 2) + \frac{1}{2} \cdot 2.40 \cdot (t - 2)^2
\]

### Step 3: Set up the equation for when the police car overtakes the speeder
The police car will overtake the speeder when their positions are equal:
\[
x_{\text{police}} = x_{\text{speeder}}
\]

So, we equate the two positions:
\[
52.78 + 26.39 \cdot (t - 2) + \frac{1}{2} \cdot 2.40 \cdot (t - 2)^2 = 31.94 \cdot t
\]

Now we solve this equation for \( t \). Let's compute that.

The time \( t \) when the police car overtakes the speeder is approximately:

\[
t \approx 8.13 \, \text{seconds}
\]

This means the police car will overtake the speeder approximately 8.13 seconds after the speeder passes it. Since this includes the initial 2.00 seconds of delay, the police car takes \( 8.13 - 2.00 = 6.13 \) seconds from the moment it starts accelerating to catch up to the speeder.

Would you like more details or have any questions about the solution?

Here are 5 follow-up questions to consider:
1. How would the result change if the police car's acceleration was higher?
2. What happens if the speeder also starts accelerating after a certain time?
3. Can you calculate the distance the police car traveled by the time it overtakes the speeder?
4. How long would it take for the police car to catch up if the speeder's speed was even higher?
5. What if the police car didn't accelerate—how far would it fall behind?

**Tip:** Always ensure that units are consistent when solving physics problems. Convert speeds to the same units (e.g., m/s) before solving equations.

An unmarked police car traveling a constant 95.0 km/h is passed by a speeder traveling 115 km/h. Precisely 2.00 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.40 m/s², how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?

Learn how to calculate the time it takes for a police car to overtake a speeder, using kinematic equations and quadratic solving. This problem covers key concepts of acceleration, constant speed, and quadratic equations. Suitable for Grades 10-12 physics students.

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