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Homework Help: Distance traveled when accelerateing

  1. Feb 12, 2010 #1
    Hey, my question is what would be the equation for figureing out distance traveled per second when accelerating from 0 at something like 12m/s2
     
  2. jcsd
  3. Feb 12, 2010 #2

    jhae2.718

    User Avatar
    Gold Member

    What else do you know? Try basic kinematics equations such as:
    [itex]
    x=x_0+v_0t+\frac{at^2}{2}
    [/itex]
    [itex]
    v_f^2=v_0^2+2a{\Delta}x
    [/itex]
     
  4. Feb 12, 2010 #3
    acceleration 12.5m/s2 accross 1800m in 12 seconds. final velocity of 150m/s

    Thanks for the reply jhae, but i cant read the equations you posted.. its just a black box. maybe because of my browser?
     
  5. Feb 12, 2010 #4
    X = Xo + Vot + 1/2 at^2

    Vf^2 = Vo^2 + 2a(Xf-Xo)
     
  6. Feb 13, 2010 #5
    Sorry for being a nuisance, but Im really new to all this, and I cant figure out how to use those equations.
    Ill lay out what I've done so far, its all been very simple up until this last part.

    Vehicle moving with uniform acceleration travels a distance of 1800m in 12 seconds, find the acceleration and distance traveled during the first and twelfth seconds.

    [tex]\Delta[/tex]v=[tex]\Delta[/tex]d/[tex]\Delta[/tex]t =150m/s
    a=[tex]\Delta[/tex]v/[tex]\Delta[/tex]t=12.5m/s^2

    Now, Im lost. My trouble is Im not sure how to use your equations.. Im not looking for the answer but maybe just a little explanation of how to use those two equations? the first one especialy
     
  7. Feb 13, 2010 #6
    Well, you need to find the distance traveled during the first second. So we need an equation that has the variables position acceleration and time which would be this...
    X = Xo + Vot + 1/2 at^2
    X=final position=what we want to find
    Xo = initial position = 0
    Vo=initial velocity=o
    t=time=1
    a=acceleration=12.5
    So if we plug in everything we get x=6.25
    Now we need to find the distance traveled in the 12th second. We can easily find that by subtracting how far the vehicle travels in 11s from how far it travels in 12s. Use the eq above.
     
  8. Feb 14, 2010 #7
    X=Xo+Vot+1/2at^2
    =0+0x12+1/2(12.5x12^2)
    =1/2 1800
    =900

    So this is saying that its postion in the 12th second is 900m? In the 12th second it should be 1800 meters, shouldnt it? Which is what we had before we divided. :confused: Im definatly doing something wrong lol
     
  9. Feb 14, 2010 #8
    I also tried figuring out the velocity and position for each second, then plugging them in under Xo and Vo but it didn't work, the number was too high too fast
     
  10. Feb 14, 2010 #9
    Is the car stationary when it starts accelerating?
     
  11. Feb 14, 2010 #10
  12. Feb 14, 2010 #11
    So, Im pretty sure in order to figure out its position during each second its as simple as

    x=at^2

    This gives me 1800m in the 12th second.
    I subtract second 12(1800m) from second 11(1512.5) and get 287.5m as the distance traveled in the 12th second.
     
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