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Distance with Initial Velocity other then 0

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A car is traveling at a straight line with an initial velocity of 14 m/s accelerates at a rate of 2.0 m/s^2 to a velocity of 24 m/s. What is the distance covered by the car in this process?

    2. Relevant equations

    d=Vot + 1/2at^2

    3. The attempt at a solution

    Now I'm not sure if you would take the final velocity of 24 m/s subtract that from the initial velocity of 14 m/s to get 10 m/s. Or do I do something totally different?

    I'm not sure how to plug the numbers into the equation. I've been looking over my notes, textbook, and power point slides but I'm confused. I tried emailing my professor hours ago but of course haven't gotten a reply. Any help you could provide me with would be greatly appreciated. Thank you!!
  2. jcsd
  3. Oct 7, 2009 #2


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    Always thought "Enter the Dragon" was your best. Anyway, mark your drawing with the x-y origin at the beginning of the acceleration, so we have a common reference point.

    That equation you picked does not contain vx(t) and vx0, can you pick another?
  4. Oct 7, 2009 #3
    This is the only problem that I was given to finish. I'm not sure what you are referring to as a drawing. Which drawing?
  5. Oct 7, 2009 #4


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    Sorry, I was not clear. I am used to drawing something that represents the problem so to better understand the problem. In this case, everything takes place on the "x-axis". I was suggesting that we declare "t=0" the moment the acceleration is applied and that the vehicle is travelling in the "+x" direction. I am familiar with 4 kinematic equations that apply to motion in a given axis, and was suggesting you look at them to find another one that would help you proceed. The equation you posted needs the "time" of the interval and the one I am hinting at would give you time.
  6. Oct 7, 2009 #5
    I just a bit ago figured out that the time is 5 seconds.

    I used V=Vo + at to find the time.

    So I used the distance formula: D= Vot + 1/2at^2

    D=(14)(5) + 1/2(2)(5)^2

    Ended up with D = 95m

    I also found a different formula which seems to be easier. It is D = 1/2 (Vf + Vi)(t)

    Plugging that in I get D = 1/2(38)(5) D = 95m

    If this isn't right I think I might just chop off my head. I've been sick with the flu and its about all the energy I have trying to figure out these damn physics problems. Anyways, thanks for all of your help.
  7. Oct 7, 2009 #6


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    Looks right.
  8. Oct 7, 2009 #7
    Thanks again for your help Lewando. I'm sure I'll be back again before the end of the quarter haha.
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