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Homework Help: Minimum stopping distance

  1. Sep 14, 2007 #1

    klm

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    The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the drivers reaction time of 0.560 s.
    what is the min. stopping distance for the same car traveling at a speed of 38.0 m/s?

    tried to use this equation v^2 - u^2 = 2 a s
    but then didnt know what to do.
     
  2. jcsd
  3. Sep 14, 2007 #2

    Doc Al

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    Staff: Mentor

    Use that equation to figure out the car's acceleration. What distance did you use? (How far has the car traveled before the brakes are applied?)
     
  4. Sep 14, 2007 #3

    klm

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    43.2m is the real distance he first travels with the velocity of 30 m/s. because i subtracted 16.8 m from the 60 m.
    so my a= 10.416
     
  5. Sep 14, 2007 #4

    Doc Al

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    Good! Now apply that--in reverse--to the second case, where the speed is 38 m/s, to find the new total stopping distance. (You'll have to calculate a new "reaction time distance".) The car's acceleration remains the same.
     
  6. Sep 14, 2007 #5

    klm

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    i dont know how to find a new reaction time distance??
    i used this formula :
    0^2 - 38^2 = 2(10.416) x
    and found x to be 69.3m but this is incorrect. can you please help me.
     
  7. Sep 14, 2007 #6

    Doc Al

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    Do it the same way as before. How far does the car move before the brakes are applied?
     
  8. Sep 14, 2007 #7

    klm

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    38 x .56?
    =21.28 m
    but what do i do with that?
     
  9. Sep 14, 2007 #8

    Doc Al

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    You add it to the other distance. Understand what happens when the person wants to stop: (1) there's a time delay due to reaction time, so the car moves some distance before the brakes are even applied; (2) once the brakes are applied, the car accelerates to a stop.

    The total stopping distance is the sum of both of these distances.
     
  10. Sep 14, 2007 #9

    klm

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    you add 21.28 to 60? to get 81.28m?
     
  11. Sep 14, 2007 #10

    Doc Al

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    No. 60m was the total stopping distance for the 30m/s case--nothing to do with the 38 m/s case!

    Add 21.28 m to the distance you calculated for the de-accelerating phase of the motion.
     
  12. Sep 14, 2007 #11

    klm

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    to the 69.3?
    69.3+21.28 = 90.58?
    sorry if i leave really quick i have to run to class in about 2 mins!
     
  13. Sep 14, 2007 #12

    Doc Al

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    Yes.
     
  14. Sep 14, 2007 #13

    klm

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    THANK YOU!!! so so much
     
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