My book has stumped me on something that I thought should be simple... perhaps I'm just not getting it.(adsbygoogle = window.adsbygoogle || []).push({});

Section - Constant Acceleration

It states the following:

"When acceleration is constant, the average acceleration and instantaneous acceleration are equal and we can write...

a = a_{avg}= (v-v_{0})/(t-0)

Here v_{0}is the velocity at time t = 0 and v is the velocity at any later time t. We can recast this equation as:

v = v_{0}+ at (Eq. 2-11)

As a check, note that this equation reduces to v = v_{0}for t = 0, as it must.As a further check, take the derivative of Eq. 2-11. Doing so yields dv/dt = a, which is the definition of a."

I'm not catching the last (bold) part. The derivative of equation 2-11 being equal to a.

When I work out the derivative (in my mind) is should work like this:

dv/dt = at

=a't + at'

=1(t) + a(1)

= t + a

My mind is leaning towards the idea that the book is trying say that dv/dt = a, at t = 0. Is that what the book is trying to tell me? Because I think I'm just confusing myself.

Any proof of this would probably help me out.

Thanks, prior!

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# Homework Help: Constant acceleration proof, dv/dt=a

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