text: # Force as a function of velocity as $t$ goes to infinity, strange result
  
  Here is the question:
  
  "A particle with mass m is given an initial velocity $$v_0$$ so that it moves in a straight line (you can consider it positive). It is subject only to a force that is inversely proportional to the square of its speed, as $$F = −c_xv^2$$.
  
  (a) Write down Newton’s 2nd law for the particle, and the corresponding differential equation for the velocity as a function of time.
  
  (b) Integrate the equation of motion to find out v(t).
  
  (c) Integrate again to find out x(t).
  
  (d) What is the total distance the particle will travel? Why? For full marks, you need to explain clearly why the answer makes physical sense. Hint: What is the limit of x(t) when t → ∞?.
  
  I solved the differential equation and got:
  
  $$v(t)=\frac{1}{\frac{ct}{m} + \frac{1}{v_0}}$$
  
  And
  
  $$x(t)=\frac{m}{c}ln(ct/m+v_0^-1)+x_0$$
  
  But as t goes to infinity, v approaches 0, suggesting the distance is finite, and x goes to infinity which is contradictory. Also, given that the object as an initial positive velocity and is subject to a retarding force, won't v quickly reach 0 and the object will stop moving? That's not reflected in the equations of motion.
  
  Ok, so here's what I'm thinking. The retarding force is proportional to the square of velocity. Eventually the velocity will be $$0 and every change in velocity will lead to a smaller and smaller acceleration that approaches 0 but never reaches it (or only reaches it at $$x=\infty$$.
  
  • what is cx here? Oct 17, 2020 at 6:14
  • @sheltonBenjamim it's some constant. Oct 17, 2020 at 6:28
  • It occurs to me, isn't this a non linear differential equation? Oct 17, 2020 at 6:29
  • This question might help you out: math.stackexchange.com/questions/3808999/… Oct 17, 2020 at 9:05
  • This is similar to the sum $\sum_n^\infty 1/n$. $1/n$ is decreasing but it's not decreasing fast enough for $S_n$ to converge, the sum is still divergent. This $\sum_n^\infty 1/{n^2}$ sum however converges to $\pi^2 /6$. Oct 17, 2020 at 19:39
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