How Long Does It Take for a Police Car to Catch a Speeder?

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To determine how long it takes for a police car to catch a speeder traveling at 38.0 m/s, the police car, which accelerates from rest at 2.29 m/s², must first reach the same speed. The time taken for the police car to match the speeder's speed can be calculated using basic kinematic equations. Once the police car reaches the speeder's speed, additional time is needed to cover the distance between them. The total time and distance covered by both vehicles can be derived from these calculations. Understanding these principles is crucial for solving the problem effectively.
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Hey. I need help on this question. A walkthrough would be great. Thanks in advance!
A speeder passes a parked police car at 38.0 m/s. The police car starts from rest with a uniform acceleration of 2.29 m/s^2.
a) How much time passes before the speeder is overtaken by policeman? b) How far does the speeder get before being overtaken by the police car?
 
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AladdinSane said:
Hey. I need help on this question. A walkthrough would be great. Thanks in advance!
A speeder passes a parked police car at 38.0 m/s. The police car starts from rest with a uniform acceleration of 2.29 m/s^2.
a) How much time passes before the speeder is overtaken by policeman? b) How far does the speeder get before being overtaken by the police car?

In order to overtake the speeder, to which displacement must the displacement of the police car be equal to? Which are the equations of diaplacement for the police car, and which ones are for the speeder?
 
confused... to which formulas are you referring, radou?
if anyone else has any input, please help...THANK YOU!
 
AladdinSane said:
confused... to which formulas are you referring, radou?
if anyone else has any input, please help...THANK YOU!

You can find the basic accleration/velocity equations here.

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Kinematics/SolvingSuggestion.html

Hint: start out by figuring out how long it will take the accelerating police car to achieve the same velocity (38ms) as the speeding car. Use this as T1. There will be a T2, it will be the time it takes the police car to travel the remaining distance to the speeder.
 
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