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- Thread starter slpnsldr
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- #2

jhae2.718

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[itex]

x=x_0+v_0t+\frac{at^2}{2}

[/itex]

[itex]

v_f^2=v_0^2+2a{\Delta}x

[/itex]

- #3

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Thanks for the reply jhae, but i cant read the equations you posted.. its just a black box. maybe because of my browser?

- #4

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X = Xo + Vot + 1/2 at^2

Vf^2 = Vo^2 + 2a(Xf-Xo)

Vf^2 = Vo^2 + 2a(Xf-Xo)

- #5

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Ill lay out what I've done so far, its all been very simple up until this last part.

Vehicle moving with uniform acceleration travels a distance of 1800m in 12 seconds, find the acceleration and distance traveled during the first and twelfth seconds.

[tex]\Delta[/tex]v=[tex]\Delta[/tex]d/[tex]\Delta[/tex]t =150m/s

a=[tex]\Delta[/tex]v/[tex]\Delta[/tex]t=12.5m/s^2

Now, Im lost. My trouble is Im not sure how to use your equations.. Im not looking for the answer but maybe just a little explanation of how to use those two equations? the first one especialy

- #6

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X = Xo + Vot + 1/2 at^2

X=final position=what we want to find

Xo = initial position = 0

Vo=initial velocity=o

t=time=1

a=acceleration=12.5

So if we plug in everything we get x=6.25

Now we need to find the distance traveled in the 12th second. We can easily find that by subtracting how far the vehicle travels in 11s from how far it travels in 12s. Use the eq above.

- #7

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=0+0x12+1/2(12.5x12^2)

=1/2 1800

=900

So this is saying that its postion in the 12th second is 900m? In the 12th second it should be 1800 meters, shouldnt it? Which is what we had before we divided. Im definatly doing something wrong lol

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- #9

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Is the car stationary when it starts accelerating?

- #10

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yes.

- #11

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x=at^2

This gives me 1800m in the 12th second.

I subtract second 12(1800m) from second 11(1512.5) and get 287.5m as the distance traveled in the 12th second.

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