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

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SUMMARY

The problem involves a police car accelerating at 5 m/s² to catch a speeding car traveling at a constant speed of 20 m/s. The key to solving the problem lies in equating the positions of both vehicles over time. By using the kinematic equation for the police car, \(d = \frac{1}{2}at^2\), and the equation for the speeding car, \(d = vt\), the time taken for the police car to catch up can be determined. The solution requires setting the distance equations equal to each other to find the time variable.

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  • Study the kinematic equations in detail, focusing on \(d = vt\) and \(d = \frac{1}{2}at^2\)
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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
 
Last edited:

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