Newton's second law integration

[learningphysics]

I thought we subbed F(v) out of the integral?

Yes. what do you get when you do that?

 

[Oblio]

t =mF(v) \int^{v}_{vo} dv

 

[learningphysics]

t =mF(v) \int^{v}_{vo} dv

before you take out F(v), you should replace it with Fo... so what do you get when you solve for v?

 

[Oblio]

t= mF(o) \int^{v}_{vo} dv

t= mF(o) (v-vo)

solve for v from here?

 

[learningphysics]

t= mF(o) \int^{v}_{vo} dv

t= mF(o) (v-vo)

solve for v from here?

Yes, one thing though, it should be m/Fo because the F(v) was in the denominator not the numerator.

 

[Oblio]

Typo. my bad

v = F(o)t / m + V(o) ?

 

[learningphysics]

Typo. my bad

v = F(o)t / m + V(o) ?

yup.

 

[Oblio]

So... why is this especially significant? They want to see some kind of comment.
Something to do with velocity dependence?

 

[learningphysics]

So... why is this especially significant? They want to see some kind of comment.
Something to do with velocity dependence?

What kind of motion is it?

v = F(o)t / m + V(o)

describe the kind of motion the object is undergoing.

 

[Oblio]

the motion would be linear i believe

 

[learningphysics]

the motion would be linear i believe

Yes, velocity is changing linearly. What about the acceleration?

 

[Oblio]

YES! your back :)
um...

increasing with time?

 

[learningphysics]

YES! your back :)
um...

:)

increasing with time?

nope.

 

[Oblio]

also constant?

 

[learningphysics]

also constant?

yes. compare:

v = v0 + at

with

v = v0 + (Fo/m)t

the equations have the same form.

 

[Oblio]

O ya!
I can say both are linear then?

 

[learningphysics]

O ya!
I can say both are linear then?

Well, linear with constant acceleration. Linear just means the object moves along a straight line.

 

[Oblio]

Yeah, understood.