How to Integrate the Motion Equation?

[WARGREYMONKKTL]

integration of Newton equation??

Newton's second law is insufficient to describe the motion of a particle. In addition, it requires a value for F, obtained by considering the particular physical entities with which the particle is interacting. For example, a typical resistive force may be modeled as a function of the velocity of the particle, for example:

\mathbf{F}_{\rm R} = - \lambda \mathbf{v}

with λ a positive constant. Once independent relations for each force acting on a particle are available, they can be substituted into Newton's second law to obtain an ordinary differential equation, which is called the equation of motion. Continuing the example, assume that friction is the only force acting on the particle. Then the equation of motion is

- \lambda \mathbf{v} = m \mathbf{a} = m {d\mathbf{v} \over dt}.

This can be integrated to obtain

\mathbf{v} = \mathbf{v}_0 e^{- \lambda t / m}
can some body show me step by step the intergration there?
thanks!

(Edited by HallsofIvy to correct LaTex.)

 

[HallsofIvy]

Newton's second law is insufficient to describe the motion of a particle. In addition, it requires a value for F, obtained by considering the particular physical entities with which the particle is interacting. For example, a typical resistive force may be modeled as a function of the velocity of the particle, for example:

\mathbf{F}_{\rm R} = - \lambda \mathbf{v}

with λ a positive constant. Once independent relations for each force acting on a particle are available, they can be substituted into Newton's second law to obtain an ordinary differential equation, which is called the equation of motion. Continuing the example, assume that friction is the only force acting on the particle. Then the equation of motion is

- \lambda \mathbf{v} = m \mathbf{a} = m {d\mathbf{v} \over dt}.

This can be integrated to obtain

\mathbf{v} = \mathbf{v}_0 e^{- \lambda t / m}
can some body show me step by step the intergration there?
thanks!

(Edited by HallsofIvy to correct LaTex.)

-\lambda \mathbf{v}= m \frac{d\mathbf{v}}{dt}
is what's called a "separable differential equation". We can "separate" the variables as
-\frac{\lambda}{m} dt= \frac{dv}{v}

Integrating both sides:

-\frac{\lamba}{m}t= ln v+ C
ln v= -\frac{\lambda}{m}t- C

v= C' e^{-\frac{\lambda}{m}t}
(C'= e^-C)

At t= 0 we have
v(0)= v_0= C' e^0= C'
so
v(t)= v_0e^{-\frac{\lambda}{m}t}

 

[arildno]

JUst a minor addition to Halls' general procedure:
It is unproblematic to generalize this if the velocity is a vector, rather than a scalar:
Method 1:
Do component-wise integration. This will always work.

Method 2:
For this particular problem, we have straightline motion, since the acceleration vector is parallell to the velocity vector.
Thus, we can always let one coordinate axis coincide with the direction of the velocity vector, i.e, transform our vector equation into a suitable scalar equation.

 

[WARGREYMONKKTL]

-\frac{\lamba}{m}t= ln v+ C
ln v= -\frac{\lambda}{m}t- C

v= C' e^{-\frac{\lambda}{m}t}
(C'= e^-C)
??
i dont' get it

??
i dont' get it
can you show how you get this?
please show me how you get it?

 

[WARGREYMONKKTL]

never mind i get it
anti-ln
thank for your replies!