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Accleration problem

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    A motorcycle has an initial velocity of 8m/s, it increases it's speed to 17m/s, covering a distance of 30m in that time.

    What is it's accleration? How long will it take to acclerate to 17m/s?

    2. Relevant equations

    As stated in the equation

    3. The attempt at a solution
    I faced this question on a quiz tonight and had not the slighest clue as to where to begin. I tried to gather a time for the change in speed over the distance, 17 - 8, / 30, which gave an answer of .3 seconds. I then tried plugging it in to d = voT + 1/2 a t^2, to check if it was correct, but it wasnt. The value obtained was far too small.

    This question really stumped me. In all the problems I have done before a time value was given. So, I had no idea how to approach it.

    Thanks for your help
  2. jcsd
  3. Oct 20, 2009 #2
    To find the acceleration here, you need to first use a kinematic equation that is independent of time, since you are not given one initially. If you use (Vf)^2-(Vi)^2=2ax, where x is the distance traveled, then you can solve for acceleration and then get the time.
  4. Oct 20, 2009 #3
    You should get a small acceleration by the way. Remember to convert minutes into seconds.
  5. Oct 20, 2009 #4
    Are there any other ways of solving the problem?

    I'm just curious because we covered the accleration chaper in the book, and it did not mention anything. It is just a matter of simple deriving it from the equation d=voT + 1/2 a t^2?
  6. Oct 20, 2009 #5
    The equation you posted is one of the kinematic equations. You use them by looking at what information is given in the problem, and then compare and match with the kinematic equations to see which one will solve for your unknown.

    The one you posted has distance as a function of time. The problem you have does not give you time, so this equation would not be the best to use when all you want to find is the acceleration.

    Yes, there are many other ways to do the problem, but using the kinematic equation that w3390 posted is by far the easiest way to go.

    The other ways include solving for time as I think you are suggesting, but to do so you'd probably need a system of two equations and perhaps make a substitution with one of those to solve for time.
  7. Oct 21, 2009 #6
    I plugged in the numbers, and it works! Definitely feels good to get the answer right. I do feel cheated though, because I never did learn this equation, but was tested on it.
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