How Can You Solve the Differential Equation a=dv/dt=-Cv^2 for v(t) and x(t)?

Click For Summary
SUMMARY

The discussion focuses on solving the differential equation a(t) = -Cv^2(t) to express velocity v(t) and position x(t) as functions of time t, given the initial condition v(0) = v0. Participants emphasize the importance of integrating the equation and suggest using the method of separation of variables to solve for v(t). The integration of v^2(t) is highlighted as a key step, with clarification needed on the indefinite integral and its implications for the acceleration a(t). The conversation concludes that proper application of integration techniques is essential for deriving the correct expressions.

PREREQUISITES
  • Understanding of differential equations, specifically first-order nonlinear equations.
  • Familiarity with integration techniques, particularly indefinite integrals.
  • Knowledge of the method of separation of variables in calculus.
  • Basic concepts of kinematics, including relationships between acceleration, velocity, and position.
NEXT STEPS
  • Learn the method of separation of variables in solving differential equations.
  • Study integration techniques for functions involving powers, such as ∫v^n dv.
  • Explore the implications of initial conditions in solving differential equations.
  • Investigate the physical interpretation of solutions in kinematics, particularly for variable acceleration.
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with differential equations and kinematic problems, particularly those interested in modeling motion under variable acceleration.

dagr8est
Messages
2
Reaction score
0
Given a(t) = -Cv^2(t) where C is a constant, express v(t) and x(t) as explicit functions of time t. Assume v(0) = v0

So I tried integrating both sides of the equation to get:
v(t)+C' = -C integral v^2(t) dt

but then I how am I supposed to integrate v^2(t) dt...

Also, I thought I could do it another way:

velocity
a(t) = -Cv^2(t)
v(t)/t = -Cv^2(t)
-1/(Ct) = v(t)

distance
-1/(Ct) = v(t)
-1/(Ct) = x(t)/t
-1/C = x(t)

Well the distance can't be constant so I guess that doesn't work either. I don't even know why this 2nd method doesn't work... It looks fine to me. Any help would be appreciated.
 
Physics news on Phys.org
You are correct in using integration to go from a(t) to an expression for v(t).

Are you just asking how to do the indefinite integral \int v^2 (t) dt ?
 
Yeah, I don't get how to integrate that. Wouldn't it be something like v^3(t)/(3a(t)) but where does the a(t) come from? It's not a constant so you just can't add it in right?
 
what??
notice that:
a=\frac{dv}{dt}=-Cv^2
how can you solve this differential equation?

hint: separation of variable!
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
30
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K