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Constant accerleration problem

  1. Sep 9, 2009 #1
    You are arguing over a cell phone while trailing an unmarked police car by 25m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0s. At the beginning of that 2.0s, the police officer begins braking sudenly at 5.0 m/s^2.
    a) What is the separation between the two cars when you attention finally returns? Suppose that you take another 0.40s to realize your danger and begin braking.
    b) If you too brake at 5.0 m/s^2, what is your speed when you hit the police car?

    for part a), i just used the equation x = x0 + v0t + at^2 / 2. a = 5, t = 2, and v0 = 0 because the the 2 cars are going in the same direction so the relative velocity of the police car is just 0. plugging those values in, i got x - x0 = 10. since the displacement is 10, and the policeman was applying the brakes, he goes back 10m so the distance between the 2 cars is 25-10 = 15m now. this was my book's answer.

    for part b) i cannot figure out what to do. i am trying to use the formula v^2 = v0^2 + 2a(x-x0). i said that x-x0 = 15, a = 5, and v0 = 12 because the brakes were applied to the police car for 2.4 seconds now so 5 * 2.4 = 12 m/s so the difference between the car's velocity and the police car's velocity at this point in time is 12. but i don't seem to get the right answer which is 94 km/h. please help.
     
  2. jcsd
  3. Sep 10, 2009 #2

    kuruman

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    First you need to repeat part (a) (except the time now is 2.4 seconds) to find the new separation between the two cars. Then you need to find how fast the police car is moving after 2.4 seconds of braking. You already know your velocity, so this will give you the relative velocity. Does the relative velocity change or does it stay the same?
     
  4. Sep 10, 2009 #3
    i get 18.56 m/s as the new speed of the police car after 2.4s. so the new relative velocity is 12 m/s. i re-did part a) using t=2.4 and i got 16.16m as the new separation. so i plugged (x-x0)=16.16, v0=12, and a=-5 into the equation v^2 = v0^2 + 2a(x-x0) but i got v^2 = -17.6. i still don't know what i am doing wrong.
     
  5. Sep 11, 2009 #4

    kuruman

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    The relative velocity is correct, but the 16.16 m separation is not. If the separation is 15 m at 2.0 seconds, it cannot be greater than that at 2.4 seconds because the cars are getting closer together. You need to redo this part.

    Assuming that you have the correct new distance, note that the relative velocity is constant throughout the braking process (Why?). So if you know the distance and you know that the relative velocity is constant, you can find the time elapsed from the time you start braking to the collision. Use that time to find your velocity just before the collision.
     
    Last edited: Sep 11, 2009
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