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Finding Final Velocity [Need Assistance]

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Question: A motorcycle traveling at 40 ft/s is given an average acceleration of 4 ft/s^(2) for 10 s. Find the final velocity.

    2. Relevant equations
    Average Velocity= [displacement][/total travel time
    ]


    v= [[x][/2] - [x][/1]][/[t][/2] -[t][/1]]


    3. The attempt at a solution

    I have done my reading on this subject (college book) and went through this in class but for some reason this question stumps me with three given variable data.The way I approached it was put the (4 ft/s^(2)) x (10 s) together which is then 40 ft/s then put the other 40 ft/s together and I would get....0? anyways can someone explain how to approach and solve. Thank You.

    40 ft/s - 40 ft/s = 0 ft/s?
     
  2. jcsd
  3. Sep 9, 2009 #2

    S_Happens

    User Avatar
    Gold Member

    I replied in your other thread before you posted this one with the formula you used.

    You are using the wrong formula and using it incorrectly. That formula calls for two positions (you don't have a final position yet) and two times, to give an average velocity for that period, which is not what you're looking for. You're looking for final velocity. I'll use you as my Latex guinea pig as I've never used it yet (and it looks like you need to learn as well).

    This is what you need to use.

    vf = v0 + a(t)

    where vf is the final velocity
    v0 is initial velocity
    a is acceleration
    t is time
     
  4. Sep 10, 2009 #3
    When I got to the Physics Forum I didn't realize I wasn't suppose to post in the Physics General Forum and when I found out yesterday probably after 5 min. of posting it up I thought I deleted it by going to my "PDF" then "unsubscribing it" which I thought I deleted my other post and then I made this post. (Sorry still new to this forum). Anways thanks for the formula because its perfect. We actually didnt get this far in class yet I beleived we stopped at "Average Acceleration". And again thanks for your help!
     
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