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Can someone start this problem for me?

  1. Sep 23, 2005 #1
    One-dimensional motion with constant acceleration

    a body moving with uniform acceleration has a velocity of 12.0cm/s when its x coordinate is 3.00 cm. If its x coordinate 2.00s later is -5.00cm, what is the magnitude of its acceleration?

    i am guessing that you use the equation for velocity in form of displacement, but im not sure at all

    some help on just picking an equation to use for this problem and starting the problem for me wud be greatly appreciated. more posts coming because im so totally lost in this chapter.
    Last edited: Sep 23, 2005
  2. jcsd
  3. Sep 23, 2005 #2


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    For uniform acceleration:
    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex] (you can get the equation by integrating)
  4. Sep 23, 2005 #3
    im still not getting it

    i knew about this equation, but i still dunt no what im doing wrong. i plugged in: (-5cm)=(3cm)+(12cm/s)+(1/2)(a)(2s)
  5. Sep 23, 2005 #4


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    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
    Solve for a:
    [tex]a = 2\frac{x-x_0-v_0t}{t^2}[/tex]
    [itex]x, x_0, v_0, t[/itex] are given.
  6. Sep 23, 2005 #5

    i got -10 cm/s^2

    but in the back of the book it says 16.

    i put x=-5 [tex]x_0[/tex]=3 [tex]v_0[/tex]=12 and t=2s

    i believe im getting [tex]v_0[/tex] wrong
  7. Sep 23, 2005 #6


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    I get (without units):
    [tex]a = 2\frac{-5-3-12*2}{2^2} = -16[/tex]
    Did you notice it's [itex]v_0t[/itex], meaning [itex]v_0[/itex] times [itex]t[/itex]?
  8. Sep 23, 2005 #7
    :surprised :surprised :surprised :surprised wow thx, i wrote my equation on my equation sheet wrong.
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