Integration of dynamics problem

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Homework Help Overview

The problem involves the dynamics of a package released from an aircraft, focusing on the effects of gravity and drag force on its motion. The subject area includes dynamics and integration of motion equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the motion equations, with some attempting to derive the velocity function over time. Questions arise regarding the integration constant and its relation to initial conditions.

Discussion Status

The discussion is active, with participants providing insights into the integration process and questioning the presence of the initial velocity in the final expression. There is no explicit consensus yet, but guidance on the importance of the integration constant has been noted.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The initial conditions and the role of the drag force are also points of discussion.

fysiikka111
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Homework Statement


A package of mass m is released from an aircraft flying horizontally with a velocity v0. In addition to gravity, a drag force acts on the package given by F=-[tex]\lambda[/tex]v. Find the i and j components of the velocity as a function of the time (t) from when the package was released.


Homework Equations


F=ma


The Attempt at a Solution


Acceleration in x-direction:
F=ma(x)
-[tex]\lambda[/tex]v(x)=ma(x)
a(x)=-[tex]\lambda[/tex]v(x)/m
dv(x)/dt = -[tex]\lambda[/tex]v(x)/m
The solution method then integrates, and gets:
v(x) = v0exp(-[tex]\lambda[/tex]t/m)
I don't know how to do the integration.
Thanks.
 
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They move all v-s to one side and t-s to the other side.
[tex] \frac{dv}{v}=-\frac{\lambda}{m} dt[/tex]

You can now easily integrate this equation and solve for v.
 
So,
dv/v = -[tex]\lambda[/tex]dt/m
ln v = -[tex]\lambda[/tex]t/m
v = exp(-[tex]\lambda[/tex]t/m)

But in the answer, where does the extra v0 come from that is before the exponent?
 
You forgot the integration constant in combination with [itex]v(t=0)=v_0[/itex].
 

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