Can anyone explain this integration for me?

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SUMMARY

The discussion revolves around the integration of a function related to motion, specifically addressing the equation m∫x''(t) dt = F_o(t) + V_o. The user initially misinterprets V_o, questioning whether it represents initial or final velocity. Upon clarification, it is established that V_o is the constant of integration, which is a standard practice in calculus. The user acknowledges the importance of including a constant of integration in their calculations.

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  • Understanding of basic calculus, specifically integration and differentiation.
  • Familiarity with Newton's second law of motion and the relationship between force, mass, and acceleration.
  • Knowledge of constants of integration in calculus.
  • Basic physics concepts related to motion and velocity.
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  • Study the concept of constants of integration in calculus.
  • Learn about Newton's laws of motion and their applications in physics.
  • Explore advanced integration techniques, including definite and indefinite integrals.
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Students studying calculus and physics, particularly those focusing on motion dynamics and integration techniques. This discussion is beneficial for anyone seeking to understand the application of integration in physical equations.

Rijad Hadzic
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Homework Statement


http://imgur.com/a/Y8NW0

Basically we start with a function of t, which was differentiated twice, that function = F_o / m

Fo is a constant force, and I assume m is mass though my book doesn't state that.

Homework Equations

The Attempt at a Solution



Integrating the function, I multiply m to the other side, because I assume m is mass and mass is a constant. so it shouldn't really matter.

since F_o is a force, that means it is Kg * m/s^2, divided by mass which is in kg, you get m/s^2 which is indeed acceleration, the second derivative of time.

Integrating, I get

m\int x^{''}(t) \, dt = \int F_o \, dt

m\int x^{''}(t) \, dt = F_o(t)

But as you see from the link I provided, the book states
m\int x^{''}(t) \, dt = F_o(t) + V_o

First off, I don't even know what V_o is. Is it inital velocity, or is it final velocity? It would make sense for it to be final velocity since m\int x^{''}(t) \, dt is literally velocity itself..

I was under the impression that if you integrate something with respect to time, and that function does not have a time variable, it would simply be that function times the variable t.

Why is it that they are adding V_o in this example?
 
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I marked it as solved. I guess I didn't read clearly enough, V_o is the constant of integration.

That makes sense now. Always end your integrals with + C, I feel dumb now..
 

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