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Finding the stopping distance of a car? Help please.

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    I am given 3 variables, Vxi (initial velocity), Tr (reaction time), Ax (acceleration.

    Now, I must make a formula for the braking distance.

    Here are the variables: Vxi = 20 m/s, Tr = 0.60 s, Ax = -4 m/s^2

    How would I construct there in a formula to get the total braking distance d = 62 m?

    2. Relevant equations



    3. The attempt at a solution

    Yes the answer is given, but I do not know the formula used, given the variables I stated above to get d = 62 m?

    Any help is much appreciated!
     
  2. jcsd
  3. Sep 23, 2011 #2
    Do you know the formulas for motion with constant acceleration?

    Method 1:

    You also know that Vx=0, since the car is going to stop.
    Using Vx=Vxi+Ax*Tb, where Tb is braking time, to find Tb, gives you the opportunity to find d using the formula:

    d=Vxi *T + (1/2)*Ax*T^2 , where T=Tr+Tb.


    Method 2:

    Use 2Ax*d=Vx^2-Vxi^2 to find distance traveled during braking, and use d=Vxi*Tr to find distance traveled during the driver's reaction time. Then find d by adding these to displacements together.

    Feel free to choose the method you feel most convenient...

    Any further questions?
     
  4. Sep 23, 2011 #3
    I s hould of mentioned that t time should not appear in the formula.

     
  5. Sep 23, 2011 #4
    Okey, I think my method 2 is somewhat easier than metod 1.
    You must take the reaction time into consideration, because it's the time before the driver pushes the brake in, and Ax starts working, therefore:

    Find distance traveled during reaction time:

    Dr=Vxi*Tr = 20m/s*0.60s =

    Find distance traveled during braking:

    [tex]2Ax *Db=Vx^2-Vxi^2[/tex] gives

    [tex]Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}[/tex]

    Stopping distance is then: d=Dr+Db.

    This will give you the 62 m you need to have...
     
    Last edited: Sep 23, 2011
  6. Sep 23, 2011 #5
    Fantastic...VERY helpfull!! Appreciate it!

     
  7. Sep 23, 2011 #6
    wait, can you please explain

    Dr and Db ?What 2 variables are you figuring out, and then subtracting them?
     
  8. Sep 23, 2011 #7
    Since there are two different parts of this movement

    Dr = distance traveled during reaction time.

    During the 0.60 s of the reaction time, the car moves with the initial velocity, Vxi all the time, then the reaction time comes to and end after 0.60 s, when the driver of the car pushes the brake in. Here acceleration is zero, and we can use v=distance/time to find distance= velocity * time.

    Db= Distance traveled during braking (that is with the brake pushed in).

    When the driver of the car has ushed the brake in, the car starts to decelerate with the acceleration of -4 m/s^2. When the car stops, the speed is 0.
    So we can solve the equation I gave you for d, here Db, to find distance traveled after the driver has applied to the brakes.

    I hope this explained it somehow, if something's still unclear, please tell me ...
     
  9. Sep 23, 2011 #8
    wow, you explained perfect! Thank you mstub

     
  10. Sep 23, 2011 #9
    Also I am not subtracting these two, I'm adding them together :)
     
  11. Sep 23, 2011 #10
    Glad to help :)

    It's always a good feeling when someone understands what you're trying to explain :D
     
  12. Sep 23, 2011 #11
    Hey, having a little trouble here.

    I get 12m for Dr

    But im not getting correct numbers for Db ? im getting 2.5...?
     
  13. Sep 23, 2011 #12
    How do you do that?

    [tex]Db=\frac {Vx^2-Vxi^2}{2Ax}=\frac {(0 m/s)^2-(20m/s)^2}{2*-4 m/s^2}=\frac {400 (m/s)^2}{8m/s^2} = 50[/tex]

    Look over your work once again.

    Sorry for the late answer, I only got an error message when trying to post: 504 gateway error...

     
  14. Sep 23, 2011 #13
    And Dr is correct :biggrin:
     
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