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Kinematics- Stopping distance

  1. Jun 14, 2008 #1
    Kinematics-- Stopping distance


    The minimum stopping distance for a car traveling at a speed of 30 m/s is 60 m, including the distance traveled during the driver's reaction time of 0.50 s. What is the minimum stopping distance for the same car traveling at a speed of 40 m/s?

    2. Relevant equations

    vf^2 = vi^2 + 2ad

    a = (delta v)/ (delta t)

    3. The attempt at a solution

    Since the reaction time is 0.5 s, the distance travelled is 40 x 0.5= 20 m. Now, to find the time it takes to decrease the velocity to 0 m/s, one needs to find the acceleration whcih can be calculated in two ways and can someone please explain to me why the first way is incorrect!

    Calculate Acceleration:
    1) use equation

    vf^2 = vi^2 + 2ad

    I use the data for the initial velocity (first scenario) 30 m/s and the stopping distance 60 m, plug it into the eequation and i get -10 m/s^2. When I use this acceleration, vf (0 m/s) and vi (40 m/s) I can calculate d=80 m, which i add to the 20 m which it takes to react and the answer I get is 100 m.

    1) However, if i use the equation, in which delta t= 2s - 0.5 s = 1.5 s. I subtracted the reaction time, the acceleration I get is a= (delta v)/(delta t) = (-40)/ (1.5) = -20 m/s^2. This acceleration is different from the acceleration calculated using the first equation. Can someone please tell me why this method is incorrect?

    Thanks a lot!
  2. jcsd
  3. Jun 14, 2008 #2


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    you've gone wrong here with your statement that delta t = 2s.
    Where did this come from? If you are attempting to calculate the stopping time during acceleration in the first part of the problem, which itself is not correct, you should also note that the stop time during acceleration in the second part is not the same.
  4. Jun 14, 2008 #3
    Oh right. That delta 2 was a stupid mistake as I divide the initial velocity by the distance to get 2 s :S which was just out of habit. Thanks for the correction! I get it now!

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