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Homework Help: Displacement with magnitude of deceleration

  1. Sep 25, 2007 #1
    You are driving your car and the traffic light ahead turns red. You apply the brakes for 3.59 s, and the velocity of the car decreases to +4.99 m/s. If the car's deceleration has a magnitude of 2.53 m/s2, what is the car's displacement during this time?

    I know Δd=Vi(Δt)+1/2at2. I am not sure how to handle the deceleration magnitude and if this is the best formula to use.
  2. jcsd
  3. Sep 25, 2007 #2


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    You can use this formula (for some reason it is less commonly mentioned or used):

    d = vf*t - (1/2)at^2

    where vf is the final velocity. since it is decelerating a = -2.53m/s^2

    or you can you use formula with vi... so first get vi using

    vf = vi + at
  4. Sep 25, 2007 #3
    so if I understand you properly:
    d= 4.99m/s * 3.59s - (.5)(-2.53m/s^2)(3.59)^2

    I get 17.914--16.30 = 34.21

    Am I using the deceleration properly? I haven't had physics in 30+ yrs
  5. Sep 25, 2007 #4


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    Yup. That's right.

    Also try calculating vi and then using

    d = vi*t + 0.5at^2

    you should get the same result that way also. that's a good way to double check your answer.
  6. Sep 25, 2007 #5
    I don't understand how to use the second formula with the numbers given - the Vf=vi+at.
  7. Sep 25, 2007 #6


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    you can get vi from that equation:

    4.99 = vi + (-2.53)(3.59)

    vi = 14.0727

    then you can use d = vi*t + (1/2)at^2
  8. Sep 25, 2007 #7


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    Then using

    d = vi*t + (1/2)at^2 = 14.0727*3.59 + (1/2)(-2.53)(3.59)^2 = 34.2175m

    So it's the same answer (only a slight difference due to rounding)

    The best way to solve this problem, is using d = vf*t - (1/2)at^2 as you did... but for some strange reason this formula isn't mentioned in physics texts very often... that's why I mentioned this other method also...
  9. Sep 25, 2007 #8
    OMG! It worked!! (14.0727)(3.59)+(.5)(-2.53)(3.59)^2 is
    50.52-16.30 is 34.21!!!!!! yipee.
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