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I can't seem to understand 1d motion.

  1. Sep 11, 2008 #1
    1. The problem statement
    I am studying for an upcoming test and I am having a lot of trouble relating velocity, acceleration, and displacement vs time.

    Given a position vs time function:
    x(t) = 6m - (8m/s)t + (1m/s^2)t^2

    I cannot for the life of me understand how to find any instantaneous velocity at any time (t), the acceleration, or any time (t) that the particle changes direction.


    2. Relevant equations

    Equations that I have been trying to use to determine the information are:
    The limit t->0 of Dx/Dt (Instantaneous velocity, change in position over change in time)
    The limit t->0 of Dv/Dt (Instantaneous acceleration, change in velocity over change in time)

    Or should I be focusing on the kinematic equation:
    X=Xi + (Vi * t) + 1/2 At^2

    3. The attempt at a solution

    I do not even know where to start. Vectors seem so much easier than this. :(



    EDIT: I am going to feel very stupid if the answer is sitting in front of me like I think it is. I just noticed that the equation given to me is in the exact form of one of the kinematic equations, minus the 1/2 in front of A, does that mean that acceleration is just 2m/s^2 (2 * 1/2 * 1m/s^2)? Likewise, is the initial velocity -8m/s? If I am right on about that, the time at which the velocity reaches 0 (changes sign) would be 4 sec?
     
    Last edited: Sep 11, 2008
  2. jcsd
  3. Sep 11, 2008 #2

    alphysicist

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    Hi jllorens,

    That's looks right to me; the acceleration is 2m/s^2 and the initial speed is -8m/s.
     
  4. Sep 11, 2008 #3

    rl.bhat

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    Yes. You are right.
     
  5. Sep 11, 2008 #4
    Thank you for the prompt replies. One more question. I have encountered two equations for average velocity, one of which seems more logical to use. If this is incorrect, I will use the other.

    Is the average velocity from t=0 to t=5:
    Using the equation for average velocity:
    Dx/Dt=Vavg

    -9 (position obtained by plugging t=5 into the given equation) minus 6 (position obtained by plugging t=0 into the given equation) divided by 5 sec (Tf) minus 0 sec (Ti) equals -3m/s? (-9-6)/(5-0) = -3

    If correct, then the average speed for the same duration is simply the absolute value of said velocity?
     
  6. Sep 11, 2008 #5

    rl.bhat

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    No. In average velocity we consider the ratio of total displacement to total time. And in average speed we consider the ratio of total distance covered to the total time.
    In the above problem they are not the same.
     
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