Police Car Overtaking Speeder: How to Calculate the Time and Distance?

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SUMMARY

The discussion focuses on calculating the time and distance for a police car to overtake a speeder. The police car travels at a constant speed of 90 km/h and accelerates at 2.00 m/s² after being passed by a speeder traveling at 130 km/h. The problem requires applying the equations of motion for constant acceleration and constant velocity to determine when the positions of both vehicles are equal. The key equations used are x = x₀ + v₀t + (1/2)at² for the police car and x = x₀ + v₀t for the speeder.

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  • Basic algebra skills to solve equations for time and distance.
  • Knowledge of initial conditions in motion problems, such as initial speeds and accelerations.
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  • Study the derivation and application of the equations of motion for constant acceleration.
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Homework Statement



A police ar traveling a constant 90 km/h is passed by a speeder travelig 130 km/h. 1.00s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s^2, how much time elapses after the police car is passed until it overtakes the speeder (assumed moving at a constant speed)



The Attempt at a Solution


This is my first encounter with these types of problems, at the level where I need to solve it without trial and error graphing.

Could someone provide a bit of guidance because I'm not sure how to do this problem.
 
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What is the position of the speeder at time t.
What is the position of the policeman at time t.
for what t are they equal.

motion with constant acceleration:
x = x_0 + v_0 t + (1/2) a t^2

motion with constant velocity
x = x_0 + v_0 t
 

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