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Homework Help: I'm A Beginner! Need help! -9th Grade Physics-

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data:

    A dragster accelerates from rest at 49 m/s^2. How fast is it going when it has traveled 325 m?

    I'm a complete beginner, I'm not really sure how to even approach this problem so please, with each step taken, provide a VERY detailed explanation of why and how you did this step! I'm so confused!
  2. jcsd
  3. Oct 4, 2012 #2
    we can't actually take you step by step through the problem here, but can try to help along the way

    do you have any formulas regarding constant acceleration?
  4. Oct 4, 2012 #3
    There is one kinematics equation that has everything you need to solve for final velocity.
  5. Oct 4, 2012 #4
    Well so far, the formulas I've learned are:

    v= Δx/t
    a= Δv/t
    Xf= X0 + V0t
    Xf = X0 + V0t + 1/2at^2
    Vf = V0 + at
    Vf^2 = V0^2 + 2aΔx
  6. Oct 4, 2012 #5
    so if you look at those formulas and look at the information you've been given, and then look at what you need to find

    which formula should you use
  7. Oct 4, 2012 #6
    Umm, well the question is asking how fast it is going...

    So I would have to use

    Vf = V0 + at
    Vf^2 = V0^2 + 2aΔx?

    Or does it not matter which one I use?
  8. Oct 4, 2012 #7
    you aren't given anything about the time, so the first one is not going to be useful

    so how would you use the equation you've chosen?
  9. Oct 4, 2012 #8
    "A dragster accelerates from rest at 49 m/s^2. How fast is it going when it has traveled 325 m?"

    So, Vf^2 = V0^2 + 2aΔx

    I guess
    a=49 m/s^2
    Δx would be 325?
    And its at rest? Would V0^2 be equal to 0 then?

    So then,

    Vf^2 = 0 + 2(49)Δ(325)?
  10. Oct 4, 2012 #9
    yes :)

    the key to these problems is listing what you know and what you don't know

    then find the formula best suited for what you've got
  11. Oct 4, 2012 #10

    Vf^2 = 31850?

    Vf = 178.47?
  12. Oct 4, 2012 #11
    looks good to me
  13. Oct 4, 2012 #12
    Thanks! :D
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