1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Car deceleration physics problem

  1. Nov 1, 2005 #1
    Hi, I need help with the following problem....

    The speed limit of a school zone is 40km/h. A Driver driving at this speed sees a child that crosses the street 13m in front of the car. He applies the brakes and desacelerates at 8m/s^2. If the reaction time of the driver is 0.50s, will the car stop before it hits the child?

    I did this:

    40km/h : 11.11m/s

    X = v^2 - V0^2/2a
    X = 0 - 11.11^2/-16
    X = 7.7m <== He drives 7.7m before the car stops.

    Now I look for the time that the car will last to stop.

    X = (v + v0/2)t
    7.7 = (0+11.11/2)t
    t = 1.4s

    Now I add 0.5s + 1.4s = 1.9s

    X = (v+v0/2)t
    X = (v + 11.11/2)1.9
    X = 10.5 M

    According to the book, it's wrong...

  2. jcsd
  3. Nov 1, 2005 #2


    User Avatar
    Science Advisor

    Try rearanging your method. You have solved for two different values that physically represent the same thing, i.e. the stopping distance.

    Try calculating the time to stop first by using
    [tex]V = V_o + a t [/tex]

    That will give you your 1.4 seconds to decelerate from the given speed to a stop.

    Now take that time (plus distance during the reaction time) and plug and chug with:
    [tex]\Delta X = V_o t + \frac{1}{2} a t^2[/tex]

    See what you come up with then.
    Last edited: Nov 2, 2005
  4. Nov 2, 2005 #3


    User Avatar
    Homework Helper

    To find the stopping distance, I would say:

    t_1 = 0.5s (reaction time)
    X = v_0*t_1 + (v^2 - v_0^2)/(2a)

    You could also do this in two steps:
    X_1 = v_0*t_1
    X_2 = (v^2 - v_0^2)/(2a)
    X_t = X_1 + X_2
  5. Nov 2, 2005 #4
    X=Vt >> X=11.11*0.5 = 5.55 which means when the driver put his leg on the brake the distance between the car and the student is 7.45 m and not 13 meter

    V^2=2ax >> x= (11.11^2)/(2*8) = 7.71 which means that the car will move 7.71 meter to stop, while the distance between the car and a child is less than this amount (7.45<7.71) so the car will hit the child
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Car deceleration physics problem
  1. Deceleration of a car (Replies: 2)

  2. Deceleration of a car. (Replies: 5)