Differentiating velocity function

[FaraDazed]

Homework Statement

Differentiate:

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

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:

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

 

[Dick]

Homework Statement

Differentiate:

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

The Attempt at a Solution

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

if its ignored then the answer is:

$a = \frac{dv}{dt} = 6t - 12$

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

 

[FaraDazed]

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

Yeah my bad, it was supposed to be 14.

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

 

[Curious3141]

Yeah my bad, it was supposed to be 14.

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

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.

 

[FaraDazed]

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?

 

[haruspex]

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?

Yes.