Solve Time to Catch Speeding Car: Police Officer Acceleration of 5 m/s2

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To determine how long it takes for a police officer to catch up with a speeding car, the problem involves a police car starting from rest and accelerating at 5 m/s² while the offending car moves at a constant speed of 20 m/s. The key is to focus on the positions of both vehicles over time rather than their velocities. The equations of motion can be set equal to each other, allowing for the calculation of the time it takes for the police car to reach the same position as the speeding car. By applying the kinematic equation for the accelerating police car and the constant speed of the offender, the solution can be derived. The discussion emphasizes the importance of correctly setting up the equations based on position rather than velocity.
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Homework Statement


A police car at rest is passed by a speeding car moving east at 20m/s. The police officer immediately starts to accelerate to the east at 5.0m/s. How long will it take the police officer to catch up with the offending car


Homework Equations





The Attempt at a Solution



VIcop=0
VFcop=?
Acop=5.0m/s2



VIoffender=20m/s


I don't really know which equation to start with because I feel as if I don't have enough variables. I want to solve for the final velocity of the cop then use that to get the time by making another equation where the final of the cop, becomes the initial, and the initial of the offender is the final... ahh I am so confused.
 
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Don't focus on the velocities, focus on the positions. The positions of the two vehicles will be the same twice: once at time zero when the driver passes the stationary cop car, and again when the cop car has caught up with driver. The speeds are not the same at either time.
 
The way I did it was by setting the equations equal to each other

So, since the car is moving 20m/s and the cop car is accelerating 5m/s^2:

vt=1/2at^2
 
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