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-   -   Kinetic motion (https://www.askmehelpdesk.com/showthread.php?t=286378)

  • Nov 29, 2008, 01:45 PM
    benbrad
    kinetic motion
    a 1.2 x 10^4 kg truck is travelling south at 22m/s

    a) what net force is required to bring the truck to a stop in 330m ?

    b) what is the cause of this net force?



    well for a) I've decided to plot in to an equation

    vf^2= vi^2 + 2ad

    0 = (484) + 2a(330)

    I'm sorry but I forgot how to solve the equation above.

    do you do 484/330 and /2 again ?
    I'm very confused on how to solve for a.
    please help
  • Dec 1, 2008, 07:30 AM
    ebaines

    You have:

    0 = 484 + 2a(30)

    The goal here is to get the "a" all by itself. First, subtract 484 from both sides:

    -484 = 2a(30)

    Now divide both sides by 2*30m:
    -484/60 = a

    Simplify:

    a = - 8.0667 m/s^2.

    Finally, double check the answer by substituting it back into the original equation to see if it works:

    0 = 484 - 2(8.06667)(30) ?

    0 = 484 - 484

    Yep, it works.

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