Acceleration and Distance problem

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SUMMARY

The discussion centers on a physics problem involving a turtle moving with constant acceleration. The turtle starts from rest and takes 10 seconds to travel 10 meters, reaching a speed of 1.2 m/s at the pine tree. The acceleration was calculated to be 0.04 m/s², and the velocity at the fence post was determined to be 0.8 m/s. To find the distance from the fence post when the turtle started, the first equation of motion was applied with the initial velocity (V1) set to 0, final velocity (V2) at 0.8 m/s, and acceleration (a) at 0.04 m/s².

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Homework Statement


A turtle is moving with constant acceleration along a straight ditch. He starts his stopwatch as he passes a fence post and notes that it takes him 10s to reach a pine tree 10m further along the ditch. As he passes the pine tree, his speed is 1.2m/s. How far was he from the fence post when he started from rest?

Homework Equations


V22=V12+2a[tex]\Delta[/tex]d
[tex]\Delta[/tex]d=V1[tex]\Delta[/tex]t+1/2a([tex]\Delta[/tex]t)2
[tex]\Delta[/tex]d=V2[tex]\Delta[/tex]t-1/2a([tex]\Delta[/tex]t)2


The Attempt at a Solution



So I found out acceleration using the third equation, which is .04 m/s2. So now I believe I have to figure out the velocity at the fence post, which comes to 0.8 m/s. I don't know what to do after that.
 
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You've already done all the work. All they are asking you is to solve for the distance the turtle traveled from rest to the fence post. Just use equation 1 with V1 = 0, V2 = 0.8 m/s, and a = .04 m/s^2.
 
Bhumble said:
You've already done all the work. All they are asking you is to solve for the distance the turtle traveled from rest to the fence post. Just use equation 1 with V1 = 0, V2 = 0.8 m/s, and a = .04 m/s^2.

Oh. Yeah I see that now thanks.
 

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