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1D Kinematics question

  1. Sep 15, 2008 #1
    Hello everyone. Im stumped here with this problem, i feel like it should be fairly simple but i cant seem to figure it out.

    1. The problem statement, all variables and given/known data

    If the acceleration for a given object is given by the function:

    a(t) = +(3 m/s^3) · t
    (Note: units are included in the eqn, so if [t]=s then [a]=m/s^2.)

    (a) Examining this function, you can tell that:
    a1)the size of the acceleration will: ?
    a2)the direction of the acceleration will be: ?

    (b) If vi (at t=0 s) is -3 m/s, find v(t). Then answer the below:
    (b1)the size of the velocity will: ?
    (b2)the direction of the velocity will be: ?

    (c) If xi (at t=0 s) is -7 m, find x(t). Then answer the below:
    c1)After 4 s have passed, What position (x) is the object at now? ___m
    c2)What is the total distance travelled (since t=0 s)? ___m

    2. Relevant equations

    I really dont know, Integration? maybe


    3. The attempt at a solution

    Okay so for part (a) I answered that the size would be "increasing" which is correct. Also the direction of the acceleration will be "in the + direction" which is aslo correct.
    So for part (b) i take the integral of the a(t) and get:

    v(t) = (3/2)t^2 - 3 m/s

    by examining this i answered that the size would be "increasing" and "in the + direction" both of my answers are wrong and i dont understand why.

    Okay so i skip to part (c) take the integral of v(t):

    x(t) = (1/2)t^3 - 3t - 7 m

    so
    x(4) = (1/2)(4^3) - 3(4) - 7 = -11

    -11 was not correct. :biggrin: so i thought it wise not to go on to part c2) lol

    and here i am stumped. Any help would be much appreciated. Thanks!
     
  2. jcsd
  3. Sep 15, 2008 #2

    alphysicist

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    Hi Spartan029,

    I think you squared the 4 instead of cubing it.
     
  4. Sep 15, 2008 #3

    lol, yea i messed up there, thanks for pointing it out.

    I figured out how to solve the problem thanks for helping me!!
     
  5. Sep 15, 2008 #4

    alphysicist

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    Sure, glad to help!
     
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