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I've attempted to resolve this problem

  1. Sep 18, 2011 #1
    1. The speed of a goods truck which has been shunted on to a level siding falls from 10km/h to 5km/h in moving a distance of 30m. If the retardation is constant how much further will the truck travel before coming to rest?



    2. I have tried to use the equation of motion v^2=U^2 + 2ax



    3. The attempt at a solution
    I tried transposing and made the equation equal to X, uh If I had the acceleration I could solve it but I'm not really sure how to find it from this question. I'm new to Physics, so excuse my dumbness. =.=
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 18, 2011 #2
    Hey there...
    You have X, which is the distance travelled. Try extracting the acceleration instead...
     
  4. Sep 18, 2011 #3
    Okay I'll try.
     
  5. Sep 18, 2011 #4
    Well this what I got a= V^2-U^2/2x = 25-100/2(30)= -1.25 and I'm not sure what to do with it.
     
  6. Sep 18, 2011 #5
    This is your acceleration.
    Now, using the formula you yourself have posted, how would we find the necessary distance to left to travel?
    Let me put it this way, your equation is:
    [itex]
    \Large
    {v_f}^2 = {v_0}^2 +2ax
    [/itex]
    Where v_f is the final velocity, and v_0 is the initial one.
    If the truck, in this case, is to stop still, what should its final speed be?
    The rest is known to you...
     
  7. Sep 18, 2011 #6
    The final velocity would be 0 and the initial 10?
     
  8. Sep 18, 2011 #7
    Very good, you can use that, though there's little point, since we're already given, that by the time it had reached 5 Km/h it has already travelled 30 m.
    Therefore, it would be good to start with the 5 Km/h, which will give you the answer directly.
    I've also noticed your slightly fumbling the units! Observe that in your calculations you plugged in 30m directly, but your velocities are in Km/h, i.e using Km as the unit of length.
    Remember that 1 Km = 1000m, and apply that as required.
     
  9. Sep 18, 2011 #8
    Oh, I will try because I was about to use another equation of motion, so that means this one is alright? I'll post again soon i'm going to attempt it again.
     
  10. Sep 18, 2011 #9
    Yes, this one is "a-okay"... just make sure your units are aright, and that you set up everything properly.
     
  11. Sep 18, 2011 #10
    Okay I tried this;

    a= 0-25000m/60m = -417 m/s^2

    displacement(x) = 0-25000/2(-417) and I got 29.9m =/ I think this is wrong, because I check my answer with the one in the back of the book it says it's 10m.
     
  12. Sep 18, 2011 #11
    You have to work in parts;
    First find "a" correctly, either in Km/h^2 or in m/s^2 if and only if you know how to convert, i.e, by multiplying by a thousand, and dividing by 3600(secs) of the velocities.
    Use the 10 and 5 kms first, by converting them to m/s and THEN plug them in; Find a in this manner.
    Set up a new equation of the same kind, where a is known but x isn't this time; Use 5 km/h and zero, (as we discussed) in m/s, and the "a" from the previous step; Find x and that's your answer!
     
  13. Sep 18, 2011 #12
    I thank you very much! I finally got it, I'm so stupid it was practically infront of me but I couldn't see it.
     
  14. Sep 18, 2011 #13
    As long as you're on the right path...
    Good luck,
    Daniel
     
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