I Review | Constants An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator, Part A If the police car accelerates uniformly at 3.00 m/s? and overtakes the speeder after accelerating for 5.00s, what was the speeder's speed? ? v = km/h Submit Request Answer
I Review | Constants An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator, Part A If the police car accelerates uniformly at 3.00 m/s? and overtakes the speeder after accelerating for 5.00s, what was the speeder's speed? ? v = km/h Submit Request Answer
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Physics Problem on Motion and Acceleration
**Scenario:**
An unmarked police car traveling at a constant speed of 95 km/h is passed by a speeder. Precisely 1.00 second after the speeder passes, the police officer begins to accelerate.
**Question (Part A):**
If the police car accelerates uniformly at a rate of \(3.00 \, \text{m/s}^2\) and overtakes the speeder after accelerating for 5.00 seconds, what was the speeder's speed?
**Calculation:**
- **Given:**
- Initial speed of the police car, \(u = 95 \, \text{km/h}\)
- Acceleration of the police car, \(a = 3.00 \, \text{m/s}^2\)
- Time taken to overtake the speeder, \(t = 5.00 \, \text{s}\)
- **Conversion:**
- Convert the initial speed of the police car to meters per second if needed for calculations:
\[
u = 95 \, \text{km/h} = 95 \times \frac{1000}{3600} \, \text{m/s}
\]
- **Calculation of speed of the speeder (v)**:
- Use the kinematic equations to find the speed at which the speeder was traveling.
- **Equation (for uniform acceleration):**
- Final velocity of the police car after acceleration can be found using:
\[
v_f = u + at
\]
- Set up the equation equating the distance traveled by both the speeder and the police car to solve for the speeder’s speed.
Enter the calculated speed of the speeder in the provided box and submit your answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ff1c888-964e-4444-aacc-2152d4151491%2Ff54afb89-c7b4-460f-92ed-64ae592e3edd%2Fdmhfb39_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem on Motion and Acceleration
**Scenario:**
An unmarked police car traveling at a constant speed of 95 km/h is passed by a speeder. Precisely 1.00 second after the speeder passes, the police officer begins to accelerate.
**Question (Part A):**
If the police car accelerates uniformly at a rate of \(3.00 \, \text{m/s}^2\) and overtakes the speeder after accelerating for 5.00 seconds, what was the speeder's speed?
**Calculation:**
- **Given:**
- Initial speed of the police car, \(u = 95 \, \text{km/h}\)
- Acceleration of the police car, \(a = 3.00 \, \text{m/s}^2\)
- Time taken to overtake the speeder, \(t = 5.00 \, \text{s}\)
- **Conversion:**
- Convert the initial speed of the police car to meters per second if needed for calculations:
\[
u = 95 \, \text{km/h} = 95 \times \frac{1000}{3600} \, \text{m/s}
\]
- **Calculation of speed of the speeder (v)**:
- Use the kinematic equations to find the speed at which the speeder was traveling.
- **Equation (for uniform acceleration):**
- Final velocity of the police car after acceleration can be found using:
\[
v_f = u + at
\]
- Set up the equation equating the distance traveled by both the speeder and the police car to solve for the speeder’s speed.
Enter the calculated speed of the speeder in the provided box and submit your answer.
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