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Car accelerates then brakes. What is its displacement?

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A car starts from rest and travels for 5.5 s with a uniform acceleration of +1.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.9 m/s2. The breaks are applied for 1.70 s.
    (a) How fast is the car going at the end of the braking period?
    4.47 m/s (I got this part)
    (b) How far has the car gone from its start?
    ?????? m


    2. Relevant equations

    1) V=Vi + at
    2) x=Vit + 1/2at^2
    3) V^2=Vi^2 + 2ax
    4) a=V/t
    5) v=x/t

    3. The attempt at a solution

    Vf^2=Vi^2+2a(x)
    4.47^2=7.7^2+2(1.4)(x)
    -14.04=X

    I know this obviously can't be the answer. I'm just confused as to how to do this because there is two different accelerations. Do i find the avg acceleration by adding them up and dividing by two or do i find the displacement before braking then subtract the displacement after braking? I'm not sure which formula to use?
     
  2. jcsd
  3. Sep 17, 2009 #2
    I would use equation 2 two times, and then adding the two results (for b)
     
  4. Sep 17, 2009 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi shizupple! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    These constant acceleration equations only work for one acceleration at a time.

    So use the first acceleration, to find a distance x1.

    Then use the second acceleration, to find a distance x2.

    The total distance will be x1 + x2. :smile:
     
  5. Sep 17, 2009 #4
    Re: Welcome to PF!

    I actually tried this too. I did:
    .5(1.4)(5.5)^2=21.175
    .5(-1.9)(1.7)^2=-2.7455

    I added them together without the - sign to get 23.9205.... wasnt correct. So i tried subtracting the -2.7455 to get 18.4295 (which is not even logical) and wasnt correct. :confused:
     
    Last edited: Sep 17, 2009
  6. Sep 17, 2009 #5
    The first part (t = 5,5 s) d = 0,5 a t^2, yields 21,175 as you stated.

    But in the second you have a velocity Vi, you have forgotten to factor that into the equation
     
  7. Sep 17, 2009 #6
    Hmm i'm kind of confused on this. Yes there is initial velocity but you dont know how long there is initial velocity. It just says the car accelerated then broke. Which to me means it instantly broke so when you do the first part of the equation you would be multiplying (7.7)(0) which would be zero? I dont see how you would figure in the initial velicity of before it was started braking?
     
  8. Sep 17, 2009 #7
    t is the same in the different equation, 1,70 s. d denotes distance from the point of origin, and at the point of origin Vi = 7,7 so it's important that you factor it in.

    Try to put Vi = 7,7 and t = 1,70 s along with a = - 1,9 m/s^2 into d = Vi t + 0,5at^2.

    The variable t is the same for both Vi t and 0,5at^2, otherwise t would have to be separated from each other using indexes (like t1 and t2)

    Also, you can see it like this: If the engine broke, the wheels would still be spinning for a while until the friction has stopped the vehicle. Just because the engine breaks, the car doesn't instantly stop.

    I hope I'm not too confusing ;)
     
  9. Sep 17, 2009 #8
    Thats a good explanation. I didnt know you had to figure in the time to the initial velocity like that. Thanks for your help Anden! I got the answer (31.5195) on my last guess!
     
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