Can someone start this problem for me?

  • Thread starter lunarskull
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  • #1
lunarskull
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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 I am 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 I am so totally lost in this chapter.
 
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Answers and Replies

  • #2
Päällikkö
Homework Helper
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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)
 
  • #3
lunarskull
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im still not getting it

i knew about this equation, but i still dunt no what I am doing wrong. i plugged in: (-5cm)=(3cm)+(12cm/s)+(1/2)(a)(2s)
 
  • #4
Päällikkö
Homework Helper
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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.
 
  • #5
lunarskull
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...

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 I am getting [tex]v_0[/tex] wrong
 
  • #6
Päällikkö
Homework Helper
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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]?
 
  • #7
lunarskull
28
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wow thx, i wrote my equation on my equation sheet wrong.
 

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