Question

I have question...
where is this (below) from?

Answer 1

\(\large\color{black}{ \displaystyle v_{\rm f}^{~~2}=v_{\rm i}^{~~2}+\Delta {\rm s} }\)

Answer 2

just need to know where this formula comes from...

Answer 3

Kinematics formula, we can derive it several ways :D, calculus/ algebra/ using graphs!

Answer 4

But essentially it comes from experimenting as all physics formulas :P

Answer 5

Can you post one derivation that you think is the easiest, please?

Answer 6

calculus maybe?

Answer 7

Haha, you would pick the tricky one! Ok well, you know that \[a = \frac{ dv }{ dt } = \frac{ d^2x }{ dt^2 }\] and \[v = \frac{ dx }{ dt }\] this is just something you should know, but I think the easiest way to derive this would be to use other constant acceleration formulas.

Answer 8

V=u+at ?

Answer 9

I mean you should know what acceleration and velocity means but for simplicity lets stick to \[a = \frac{ v-v_0 }{ t_2-t_1 }\] for motion along a straight line we then have \[a_x = \frac{ v_x-v_{0x} }{ t-0 }\] where we start at a initial time 0, I don't know if you really want to go really in depth with calculus mhm, as it requires integral so I think this will suffice. Note we get then \[v_x = v_0x+a_xt\] and here we have it, we got one of the equations!

Answer 10

I am fine with any integrals, as long as the step sequence is not too complicated. I have been integrating for long enough:)

Answer 11

I mean this really doesn't need it at this point, especially since we're deriving motion in a straight line, and it's much easier to visualize

Answer 12

Ok, so how would I derive this?

(The \(\large\color{black}{ \displaystyle v_{\rm f}^{~~2}=v_{\rm i}^{~~2}+\Delta {\rm s} }\) )

Answer 13

Essentially if you notice that this is just position - time, velocity - time graphs

Answer 14

I think you mean \[v_f^2=v_i^2 = 2 a \Delta s\]

Answer 15

yes I did...

Answer 16

\(\large\color{black}{ \displaystyle v_{\rm f}^{~~2}=v_{\rm i}^{~~2}+2a\Delta {\rm s} }\)

Answer 17

Can we assume we have all the other other equations as it would take quite some time to get there then..

Answer 18

I know the equations, I just need some place to get it from.

Answer 19

Some proof of this particular formula, if there is.

Answer 20

As I mentioned above it comes from motion with constant acceleration which is from experimenting, but I guess then lets use some calculus as it's the quickest I think haha! Starting with \(a = \frac{ dv }{ dt }\) \[\huge \int\limits_{0}^{t} a dt = \int\limits_{v_0}^{v_f}dv \implies at=v_f-v_i \]\[\huge \implies v_f = v_i+at\] there's one of the equations, now we can use our definitions of velocity to get \[\huge \frac{ dx }{ dt } v_0 +at \implies \int\limits_{x_0}^{x} dx = \int\limits_{0}^{t} (v_0+at)dt\]
\[\huge \implies x-x_0 = v_0t+\frac{ 1 }{ 2 }at^2\]\[\huge \implies \Delta x = v_0t+1/2at^2\]

Answer 21

For the equation you're looking for then we have \[\huge a = v \frac{ dx }{ dt } \implies \int\limits_{x_0}^{x} a dx = \int\limits_{v_0}^{v} v dv\] \[\huge \implies v_f^2 = v_0^2+2a \Delta x\]

Answer 22

I used chain rule for a = v dx/dt