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Can someone start this problem for me?

  • Thread starter lunarskull
  • Start date
  • #1
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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 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:

Answers and Replies

  • #2
Päällikkö
Homework Helper
515
10
For uniform acceleration:
[tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex] (you can get the equation by integrating)
 
  • #3
28
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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)
 
  • #4
Päällikkö
Homework Helper
515
10
[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
28
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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 im getting [tex]v_0[/tex] wrong
 
  • #6
Päällikkö
Homework Helper
515
10
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
28
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:surprised :surprised :surprised :surprised wow thx, i wrote my equation on my equation sheet wrong.
 

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