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Homework Help: Kinematic problems !

  1. Sep 22, 2003 #1
    Kinematic problems!!!!!!

    hi there,here is my problem

    The police observe that the skid marks of a stopping car are 200m long. Assuming the car decelerated at a constant rate of 20m/s^2,skidding all the way, how fast was the car going when the brakes were applied?

    i think we can solve the problem by using the physics equation v^2=u^2+(2*(a*s)).......and
    i try to use integration but the time interval isn't given so i am just looking for an alternative way to solve this problem.........by the way ,this is a calculus assignment.

  2. jcsd
  3. Sep 22, 2003 #2
    Im getting Vi as 88.5 m/s
  4. Sep 22, 2003 #3
    yes i got 89 something too.......but i just wonder if this problem can be done by integration?

    but thx anyway
    Last edited: Sep 22, 2003
  5. Sep 22, 2003 #4


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    Last edited by a moderator: Apr 20, 2017
  6. Sep 22, 2003 #5
    thx integral it really help! :) thx a lot
  7. Sep 23, 2003 #6


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    Because this is a constant acceleration problem (so that the velocity function is linear), you can use the "average" velocity.

    Let v be the initial speed. Of course, the ending speed is 0 so the "average" speed, the average of those two numbers, is v/2.

    Letting t be the (unknown) time required to stop, the distance is
    (v/2)t= 200 m.

    And we do know the time- in a sense. At an acceleration of -20 m/s^2, to go from v to 0 requires (since 0= v- 20t) v/20 sec.

    Putting that into (v/2)t= 200 gives v^2/40= 200 or v^2= 8000.

    v= [sqrt](8000)= [sqrt](1600*5)= 40[sqrt](5)= 89.4 m/s

    If you're not really sure about that "average" speed, here's the standard way:

    With acceleration a, the distance travelled is d= (a/2)t^2+ vt.
    (You could say we use integration to get that.)

    In this problem, -10t^2+ vt= 200.

    Again, since the time taken to slow from v to 0 with acceleration
    -20 m/s^2 is v/20 sec., we have (-10)v^2/400+ v^2/20= 200.
    That is, -v^2/40+ v^2/20= v^2/40= 200 so v^2= 8000 again.
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