Kinematics equations: Prove that a=vdv/ds

[Crystal037]

Homework Statement

Prove a=vdv/ds

Relevant Equations

V=dx/dt
a=dv/dt

We know that v=v(x, t) i.e. V is a function of time as well as space coordinate here I have only taken 1D motion for simplicity. By chain rule a=dv/dt= Dv/Dx*dx/dt +Dv/Dt*dt/dt where D /Dx represent partial differentiation along x coordinate.
So, a=Dv/Dx*dx/dt +Dv/Dt
Here we say that Dv/Dt=0 since we have taken displacement as constant
So a=Dv/Dx*dx/dt = Dv/Dx*v
But I am not getting what is v since velocity is constantly changing. How can we have a constant v

 

[archaic]

$\vec a=\frac{d\vec v}{dt}=\frac{d\vec v}{ds}\frac{ds}{dt}$ and $ds=|\vec v|dt$

 

[haruspex]

You are mixing up two situations.

The equation a=vdv/ds applies for a particle moving as a function of time. x=x(t), v=v(t). Yes, you can write v as a function of x, but only on the basis that the relation x=x(t) can be inverted. E.g. if x=f(t) and v=g(t) then t=f^-1(x), so v=g(f^-1(x)).

The equation $\frac d{dz}f(x, y)=\frac{\partial f}{\partial x}\frac{dx}{dz}+\frac{\partial f}{\partial y}\frac{dy}{dz}$ applies where f is a function of two independent variables, x and y.
If you try to apply it to the particle trajectory equation as $\frac d{dt}v(x, t)=\frac{\partial v}{\partial x}\frac{dx}{dt}+\frac{\partial v}{\partial t}\frac{dt}{dt}$, the two partial derivatives are meaningless since neither x nor t can change without the other.

 

[Crystal037]

But then what is v vector. It's the velocity at what point since the velocity keeps changing

$\vec a=\frac{d\vec v}{dt}=\frac{d\vec v}{ds}\frac{ds}{dt}$ and $ds=|\vec v|dt$

 

[archaic]

But then what is v vector. It's the velocity at what point since the velocity keeps changing

it is $\vec v(t)$ of a particle at $\int\vec v(t)dt+\vec x_0$

 

[Crystal037]

So v(t) is the instantaneous velocity at t which corresponds to the coordinate
i.e. Integral of v(t) w.r.t to time +x

 

[archaic]

So v(t) is the instantaneous velocity at t which corresponds to the coordinate
i.e. Integral of v(t) w.r.t to time +x

yes but we consider the integration constant to be 0

 

[Crystal037]

But v isn't constant

 

[archaic]

But v isn't constant

yes i meant when you do then integral of v you will have x + c, that c is taken to be the initial position. but since i put the initial position vector outside, i mentioned that c shouldn't be considered.

 

[archaic]

But v isn't constant

I should have written it in this form $f(t)=(\int_{t_0}^t f'(\tau)d\tau)+f(t_0)=(f(t)-f(t_0))+f(t_0)$, which is basically the fundamental theorem of calculus. Apply it to your vector knowing that $\vec v(t)=x'(t)\hat \imath+y'(t)\hat \jmath+z'(t)\hat k$