# Constant acceleration proof, dv/dt=a

1. Apr 18, 2010

### Beamsbox

My book has stumped me on something that I thought should be simple... perhaps I'm just not getting it.

Section - Constant Acceleration

It states the following:

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

a = aavg = (v-v0)/(t-0)

Here v0 is the velocity at time t = 0 and v is the velocity at any later time t. We can recast this equation as:

v = v0 + at (Eq. 2-11)

As a check, note that this equation reduces to v = v0 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!

2. Apr 18, 2010

### D H

Staff Emeritus
What is the derivative of f(t)=2 with respect to t? The derivative of f(t)=c, where c is some constant? Here, a is just another constant.

3. Apr 18, 2010

### AtticusFinch

Yeah I'm not really sure why you think this:

dv/dt = at
=a't + at'
=1(t) + a(1)
= t + a

works.

Here's how you should see it. Equation 2-11 shows v as a function of time with initial velocity v0. In other words,

v(t) = v = v0 + at

Take the first derivative with respect to time. v0 is a constant and derivative of a constant is zero. a is also a constant and the derivative of a constant times a variable is the constant itself. So...

v'(t) = a

4. Apr 18, 2010

### Beamsbox

That did it, I was thinking of at as being the product of two variables... instead of say, dv/dt = 6t = 6 for example... now THAT I can understand.

Thanks for the help, it's been a few years...

5. Apr 18, 2010

### AtticusFinch

Ah I see so you were using the product rule to get = t + a

Well, good that you understand now :)

6. Apr 18, 2010

### Beamsbox

Funny actually, now that I think about it, I'm in the CONSTANT ACCELERATION section... CONSTANT... doh!

Thanks again!