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Police car's acceleration physics problem

  • Thread starter physicsss
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  • #1
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A police car traveling a constant 95 km/h is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time passes after the speeder passes before the police car overtakes the speeder (assumed moving at constant speed)?

I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation

x = x + v*t + 1/2 * a *t^2
40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2

t=15, but that's wrong....any ideas?
 

Answers and Replies

  • #2
Tide
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[tex]95t+\frac{1}{2}at^2=140(t+1)[/tex]

... where a is the acceleration with appropriate units, of course! :-)
 
  • #3
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I still get t=15 with that equation. :confused:
 
  • #4
Tide
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You did? I got 6.3 minutes if I did my arithmetic right - seems too long though! Check to see if you converted the units on your acceleration correctly.
 
  • #5
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well, since the final answer is in seconds, I converted 140 km/h and 95km/h to m/s. So

95km/h * 1h/60min * 1min/60s * 1000m/1km= 26.389 m/s and
140km/h * 1h/60min * 1min/60s * 1000m/1km= 38.889 m/s.

so 26.389*t+1/2*2.0*t^2=38.889(t+1)
and t=15 s.


:bugeye:
 
  • #6
Tide
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Oh! The t+1 in my original equation was incorrect. I was mixing seconds with hours. When I correct for that I get something closer to what you have (15.9s).

How do you know your answer is wrong?
 
  • #7
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The hw is online and when I entered 15s and submitted, it said it was incorrect.
 
  • #8
Chronos
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You are getting distance, not acceleration, from that formula. The original formula Tide gave is correct. You need only solve for time. Keep trying.
 
Last edited:
  • #9
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Chronos, can you clarify what you said?
 
  • #10
Doc Al
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Tide said:
[tex]95t+\frac{1}{2}at^2=140(t+1)[/tex]
This isn't quite right for several reasons.

Measuring everything from the moment the speeder passes the police car (at t = 0):
Distance police car travels: [itex]x_p = v_p t + \frac{1}{2}a(t-1)^2[/itex], note that acceleration only begins at t = 1 second, [itex]v_p[/itex] is the initial speed of the police car in m/s.
Distance speeder travels: [itex]x_s = v_s t[/itex], [itex]v_s[/itex] is the speed of the speeder in m/s.

So, when the police car overtakes the speeder:
[tex]v_p t + \frac{1}{2}a(t-1)^2 = v_s t[/tex]
All times measured in seconds. Now plug in your values and solve for t.
 
  • #11
Tide
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Thanks, Al, I stand corrected!
 
  • #12
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I got 14.4 seconds, is that correct?
 

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