Re: Physics Homework Help, Please

In summary, the question involves an unmarked police car traveling at a constant speed of 80km/h being passed by a speeder traveling at 135km/h. After 3.00s, the police car accelerates at a rate of 1.50m/s2. The problem asks for the amount of time it takes for the police car to catch up to the speeder, assuming both cars maintain constant speed. The solution involves setting the position of the police car and the time at which the speeder passes it as 0, and then using equations for the distance traveled by each car to find the time when their distances are equal.
  • #1
DrAnteater
1
0

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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  • #2
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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