Why is the slope of a V2 Vs. X Graph 2a?

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The discussion centers on understanding why the slope of the V2 versus X graph is 2a. The key equations referenced include vf2 = vi2 + 2ax, which shows the relationship between final velocity squared, initial velocity squared, acceleration, and displacement. The confusion arises from where the factor of 2 comes from in this equation. By manipulating the equation for velocity, vf = vi + at, and squaring it, the derivation reveals the 2a term in relation to displacement. This clarification helps solidify the connection between the graph's slope and the underlying physics principles.
YungEggy
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I just need to know what equations and everything is used to explain why the slope of V2 Vs. X graph is 2a. I understand why V vs. T is a and X vs T2 is 1/2a.
Thanks!
 
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a = v/t (velocity vs. time)

x = (1/2) a t2 -> x / t2 = (1/2) a (x vs. t2 is a/2)

vf2 = vi2 + 2 a x

If vi2 = 0, you can rearrange the above equation to get the slope of the vf2 vs. x graph similar to the way you got the slope of the x / t2 graph.
 
But the only problem I have is where did the 2 come from? I understand how to arrange it to get the slope of a postion time graph.
 
YungEggy said:
But the only problem I have is where did the 2 come from?

Which 2? vf2 = vi2 + 2 a x This one?

Multiply the equation vf = vi + at with itself
vf2 = vi2 + a2t2 + 2atvi
vf2 = vi2 + 2a( vit + at2/2)
vf2 = vi2 + 2ax
 

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