Differentiating velocity function

1. Dec 7, 2012

1. The problem statement, all variables and given/known data

Differentiate:

$v = 3t^2 - 14t + 8$

3. The attempt at a solution

I am not sure if you do anything with the constant "+ 8" is it ignored when differentiating?

if its ignored then the answer is:

$a = \frac{dv}{dt} = 6t - 14$

Last edited: Dec 7, 2012
2. Dec 7, 2012

Dick

Oh, it is not. It's 6t-14. Isn't it? Is that just a typo?

3. Dec 7, 2012

Yeah my bad, it was supposed to be 14.

So is the +8 ignored then, if so, why?

4. Dec 7, 2012

Curious3141

It's not ignored, it's just that the derivative of a constant (8 in this case) is zero.

Think about it in the physical context. The derivative of velocity with respect to time is acceleration. Acceleration measures the rate of change of velocity, so velocity has to vary for there to be a non-zero acceleration. The terms that are dependent on t are varying, so they affect the acceleration. But the '+8' is just a constant add-on velocity. Since this doesn't change, it does not affect the acceleration. So it's correct that the term vanishes when you differentiate.

5. Dec 7, 2012

Thanks for explaining, it makes sense now.

Can you just confirm if I am doing a major *** up please?

a > v > s

to get from v to a you differentiate, and to get from v to s you integrate?

i.e. to go left you differentiate and to go right you integrate?

6. Dec 7, 2012

Yes.