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Equations for Accelerated Motion Problem 2

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data
    A car traveling at 60 mi/h stops in 6 seconds. How far does it travel during this time?

    2. Relevant equations
    a=Δv/Δt
    x=xinitial+vinitialt+1/2(a)t^2


    3. The attempt at a solution

    First I changed units into the standard ones:
    60 mi/h *1609.34 m/1 mi * 1 h/3600 s = 26.8223 (the 3 seemed to be repeating.)

    Then I used a=Δv/Δt, assuming the Vinitial was zero:
    26.8223/6
    a≈4.47 m/s^2

    Finally, I plugged my numbers into this equation:
    x=xinitial+vinitialt+1/2(a)t^2

    and got 80.46 m.

    Would someone please check that this is correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 21, 2013 #2
    That assumption isn't correct. The given velocity in the question is the initial velocity. And it stops after 6 seconds. Acceleration would be negative. And it seems you did take that into account in the final equation, otherwise the answer wouldn't be correct. Yes, the answer is correct. But be careful for the values you choose and assumptions you make, acceleration is negative anyhow in this case.

    Edit: if you take ##a## to be positive, the final equation would have a minus sign.

    ##x=x_{initial}+v_{initial}.t-\frac{1}{2}.a.t^2##
     
  4. Oct 21, 2013 #3
    Okay, thank you! I didn't take the negative acceleration into consideration at all, so thanks for pointing it out! I also used vintial twice when I meant to say one was vfinal. Oh! I got it backwards, the initial velocity was the given and the final velocity was zero. Thanks again!!
     
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