Re: Physics Homework Help, Please

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SUMMARY

The discussion revolves around a physics problem involving an unmarked police car and a speeder. The police car travels at a constant speed of 80 km/h and accelerates at 1.50 m/s² after a 3-second delay, while the speeder maintains a constant speed of 135 km/h. The key to solving the problem lies in equating the distances traveled by both vehicles to determine the time it takes for the police car to overtake the speeder. The user struggles with the equation setup, specifically in relating the distances and times correctly.

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  • Understanding of kinematic equations in physics
  • Knowledge of constant acceleration and uniform motion
  • Ability to convert units (e.g., km/h to m/s)
  • Familiarity with algebraic manipulation of equations
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  • Study kinematic equations for uniformly accelerated motion
  • Learn how to convert speeds from km/h to m/s for calculations
  • Practice solving problems involving relative motion and acceleration
  • Explore graphical representations of motion to visualize distance vs. time
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Students studying physics, particularly those focusing on kinematics and motion problems, as well as educators seeking to clarify concepts related to acceleration and relative speed.

DrAnteater
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Homework Statement



An unmarked police car traveling a constant 80km/h is passed by a speeder traveling 135km/h.

Precisely 3.00s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.50m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?


~I would sincerely appreciate help with this question. It's the only problem I'm having a difficult time understanding.


The attempt at a solution

I keep getting stuck at: 37.5t = 66.66 +22.22t - 66.66 + 1/2a(t-3)^2
Not sure if I'm even approaching it right.
 
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DrAnteater said:

Homework Statement



An unmarked police car traveling a constant 80km/h is passed by a speeder traveling 135km/h.

Precisely 3.00s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 1.50m/s2 , how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed)?


~I would sincerely appreciate help with this question. It's the only problem I'm having a difficult time understanding.


The attempt at a solution

I keep getting stuck at: 37.5t = 66.66 +22.22t - 66.66 + 1/2a(t-3)^2
Not sure if I'm even approaching it right.

Let's just set the position of the police car when the speeder passes it as ##x = 0##, and the time at which this happens ##t = 0##. So if we're looking for the time at which the police car catches up to the speeder, we're looking for the time when the distance traveled by each car is equal, right? So what are the equations for the distance traveled by each car?
 

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