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$\vec{s}$ is an arbitrary position vector, which can change with time.

$\vec{v} = \dfrac{\Delta \vec{s}}{\Delta t}$ is the velocity vector.

$\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t}$ is the acceleration vector, and it readily follows that

$\vec{v_f}=\vec{v_i}+\vec{a}\Delta t$

$\dfrac{\Delta \vec{s}}{\Delta t}=\vec{v_i}+\vec{a}\Delta t$

$\Delta \vec{s}=\vec{v_i}\Delta t+\vec{a}(\Delta t)^2$

However this differs from the accepted $\Delta x=v_0t+\frac{1}{2}at^2$

What's wrong with my reasoning and how can I get to the correct answer using purely vectors?

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1 Answer 1

Your second to last step is the statement $$\frac{\Delta{\bf s}}{\Delta t}={\bf v}_f$$ which doesn't make a lot of sense. One could argue that it makes more sense to use the velocity at the middle of the time interval $\Delta t$, in which case you get the desired result.

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I don't understand why it doesn't make sense, though, because I'm simply substituting in definitions. Is there any sort of mathematical principle that makes it clear why I would need to use the velocity in the middle of the interval? –  user1641398 Jan 20 '13 at 22:17
    
The velocity is changing in that time interval. There is no obvious reason to use the middle of the interval, but it undoubtedly makes more sense than the velocity at the beginning or end of the interval. To make a long story short, you should do this with calculus instead. –  Jonathan Jan 20 '13 at 22:23

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