Like friction, drag force opposes the motion of a particle in a fluid; however, drag force depends on the particle's velocity. Find the expression for the particle's velocity v(x) as a function of position at any point x in a fluid whose drag force is expressed as Fdrag = kmv where k is a constant, m is the mass of the particle and v is its velocity. Assume that the particle is constrained to move in the x-axis only with an initial velocity v0. Solution: The net force along the x-axis is: ΣF = -F = m then: -mv = m Since acceleration is the first time derivative of velocity a = dv
Like friction, drag force opposes the motion of a particle in a fluid; however, drag force depends on the particle's velocity. Find the expression for the particle's velocity v(x) as a function of position at any point x in a fluid whose drag force is expressed as
Fdrag = kmv
where k is a constant, m is the mass of the particle and v is its velocity. Assume that the particle is constrained to move in the x-axis only with an initial velocity v0.
Solution:
The net force along the x-axis is:
ΣF = -F = m
then:
-mv = m
Since acceleration is the first time derivative of velocity a = dv/dt,
-mv = m
We can eliminate time dt by expressing, the velocity on the left side of the equation as v = dx/dt. Manipulating the variables and simplifying, we arrive at the following expression
/ = -k
"Isolating" the infinitesimal velocity dx and integrating with respect to dx, we arrive at the following:
= v0 -
which shows that velocity decreases in a linear manner.

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