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Super congfused about finding distance with constant acceleration

  1. Sep 3, 2012 #1
    *confused

    A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?

    So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but this is wrong. I was also told to try drawing a graph of velocity vs time....but how can I graph numbers such as these (high and low #'s with decimals) If I cant graph it I cant find the slope or y intercept....so I'm lost
     
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2
    The car is accelerating at a constant rate (if I understand the problem correctly, it's wording is strange), so its velocity is constantly changing.

    Do you know any kinematic equations for motion under constant acceleration?
     
  4. Sep 3, 2012 #3

    PhanthomJay

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  5. Sep 3, 2012 #4
    In class we've used these: v=v0+at, x=x0+v0t+1/2at^2, and v^2=V0+2a(x-x0)
    but from what I see I have an intitial velocity of 0, a distance and starting point which would be x0=0 and x=100m then I have two times..... I'm having trouble fitting this all in..
     
  6. Sep 4, 2012 #5

    CAF123

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    Start by using [itex] x - x_o = v_{o}t + \frac{1}{2}at^2 [/itex] to find this acceleration, which is a constant value.
     
  7. Sep 4, 2012 #6
    Can I use the eqtn v=v0+at instead with the initial velocity being zero, velocity being 100m/6.04s and t=6.04s? and solving for a?

    and them plug it into the x-x0 eqtn?
     
    Last edited: Sep 4, 2012
  8. Sep 4, 2012 #7

    lewando

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    No. The v in that equation is final velocity. You would be using average velocity. Do what CAF123 suggests.
     
  9. Sep 4, 2012 #8
    Ok so when I solve for a the eqtn is a=2x/t^2. Then I plug a back into the eqtn but I use the second given time.... Am I on the right track?
     
  10. Sep 4, 2012 #9

    CAF123

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    Yes!
     
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