How Does Integrating Modified Newton's Law Lead to the Velocity Formula V(t)?

[bhsmith]

Homework Statement

Starting from the modified Newton's Law

(dp(rel))/dt=F

with a constant Force F, and assuming that the particle starts with v=0 at time t=0, show that the velocity at time t is given by

V(t)=c [(Ft/mc)/(1+ Ft/mc)]

Homework Equations

The Attempt at a Solution

I know that I can integrate both sides of the equation with respect to time and solve, but I'm stuck on how to start that off. Any help would be appreciated!

 

[rpf_rr]

Integrate, you find p=Ft, substitute p=mv/sqrt(1-v^2/c^2), some arithmetics and you fininshed, your solution is wrong, is valid for v^2

 

[bhsmith]

I figured that one out too. But that equation for v(t) is stated in the problem. I'm thinking it might be different because it is supposed to be a "modified" Newton's Law for relativity instead of the classical equation P(class)=mv

 

[rpf_rr]

my result is correct for relativity (at least for special as far i know), it is even reported in my textbook

 

[paranormal]

I would approach this problem by first understanding the context and assumptions of the given information. It appears that we are dealing with a particle moving under the influence of a constant force, and we are asked to find the velocity of the particle at a given time t.

To start off, let's recall the definition of acceleration from Newton's second law: a = F/m, where a is the acceleration, F is the force, and m is the mass of the particle. We can rearrange this equation to solve for the force: F = ma.

Now, let's integrate both sides of the modified Newton's Law equation with respect to time:

∫(dp(rel))/dt dt = ∫F dt

The left side of the equation becomes the change in momentum (p) of the particle, and the right side becomes the force (F) multiplied by the change in time (t). We can also rewrite the change in momentum as p = m * v, where v is the velocity of the particle.

Substituting these values into the equation, we get:

∫m * dv/dt dt = ∫F dt

We can now use the chain rule to rewrite dv/dt as dv/dx * dx/dt, where x represents the position of the particle. This allows us to change the variables in the integral to x and v:

∫m * dv/dx * dx/dt dt = ∫F dt

Next, we can use the definition of acceleration (a = dv/dt) to replace dv/dt in the integral:

∫m * a * dx/dt dt = ∫F dt

Using the definition of work (W = F * x) and the assumption that the particle starts with v = 0 at t = 0, we can rewrite the right side of the equation as W = F * x = ∫F dt.

This gives us the following equation:

∫m * a * dx/dt dt = W

Integrating both sides with respect to time, we get:

∫m * a * dx = ∫W dt

We can now use the definition of acceleration (a = F/m) and the assumption that the particle starts with v = 0 at t = 0 to simplify the left side of the equation:

∫F * dx = ∫W dt

Using the