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Reaction time and acceleration.

  1. Aug 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Part A. A motorist travels at 10 m/s when he sees a deer in the road 45m ahead. if maximum negative acceleration is -7m/s^2, what is maximum reaction time (delta t) of the motorist that will allow him to avoid hitting the deer? Answer in seconds please.
    So Initial Velocity is 10
    Acceleration is 7.
    Final Velocity is 0 (I think you can deduce that because the motorist wants to come to a stop, which would be 0?)
    Xo=0 (I'm not sure how, but that's what I think, can somebody explain?)

    Part B:
    If his reaction time is 3.97461s, how fast will he be traveling when he reaches the deer?


    2. Relevant equations
    This hw assignment was given on the three motion at constant acceleration equations:
    V=Vo +at
    X=Xo+Vot +1/2(a)t
    V^2=Vo^2+2a(delta x)


    3. The attempt at a solution
    I'm not sure which of those equations to use or if I should use past equations I have learned such as speed or acceleration or velocity. I have tried all three, but when I check my answer it is always wrong. Since I am not able to get Part A, I cannot get Part B. Can somebody help me out.
     
  2. jcsd
  3. Aug 29, 2009 #2
    You have all the things right, except for acceleration, which is -7
    Ok, a couple of hints..:

    How long would it normally take him to hit the deer if he didn't stop?
    How long does it take him to stop?
    What does this tell you about the maximum reaction time?

    It makes things easier. X now becomes not only the final position, but also the distance. There's also less variables to play with, which is obvious preferable.


    Have a go at part A now, and if you get the answer, part B. Do you know what the answer is?
     
  4. Aug 29, 2009 #3
    Sorry I meant -7 for acceleration:
    The answer is 3.786.
    How long does it take him to stop? about 1.429 seconds because 0=10 + -7t
    t= 10/7 or 1.429 seconds.
    I think I figured it out:
    If it takes him 10/7 seconds to stop, I could find how much the driver would move using the second equation right? So I tried it out: X= 0 + 10(10/7) + .5(-7)(100/49)
    x=7.143
    So if the driver pressed the brakes he would still move a total of 7.143 meters. So then if the deer is 45 meters away, he has a distance of 37.857 meters before he has to stop.

    So I used the second equation again:
    37.857 = 10(t)+ 1/2(0)(t)
    t= 3.786

    I did not put 10/7 for t because that was the time to stop, I want to know the time he can travel before he stops. Also, for a I used 0 because he was moving at constant speed of 10 m/s.

    Is my explanation right or was I lucky when I got the answer.
     
  5. Aug 29, 2009 #4
    That is the answer for part a. Sorry I forgot to mention that.
     
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