Find the particle's haricontal position a() and velocity va) at any point in a fluid whose drag force is expressed as where, k is a constant, mis the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity vo. Solution: a) To salve for the position as a function of time x(), we construct the net force in the x-axis as

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Chapter1: Units, Trigonometry. And Vectors
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Find the particle's horizontal position x(1) and velocity v(x) at any point in a fluid whase drag force is expressed as
Farag
- kmy
where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity vo.
Solution:
a) To salve for the position as a function of time x(1), we construct the net force in the x-axis as
EF-F
m
Then:
v- m
since:
a- dy/di
Lhen
by integrating, we obtain the following expression:
Further, employing the rules of integration results to the following expression for position as a function of tLime
x- (vo
as t+ 0, the positian becomes
x- vo/k
b) To solve for the velocity as a function of position v(x), we construct the net force in the x-axis as follows
F-
- m
Then:
-mi
y- m
since:
a- dy/dt
Lhen
-mi
y- m
We can eliminate time by expressing, the velacity on the left side of the equation as
v- dx/dt
Then, we arrive at the following expression
By integrating and applying the limits, we arrive at the following
-vo
which, sows that velacity decreases in a linear maner.
Transcribed Image Text:Problem Find the particle's horizontal position x(1) and velocity v(x) at any point in a fluid whase drag force is expressed as Farag - kmy where, k is a constant, m is the mass of the particle and v is its velocity. Consider that the particle is initially traveling with a velocity vo. Solution: a) To salve for the position as a function of time x(1), we construct the net force in the x-axis as EF-F m Then: v- m since: a- dy/di Lhen by integrating, we obtain the following expression: Further, employing the rules of integration results to the following expression for position as a function of tLime x- (vo as t+ 0, the positian becomes x- vo/k b) To solve for the velocity as a function of position v(x), we construct the net force in the x-axis as follows F- - m Then: -mi y- m since: a- dy/dt Lhen -mi y- m We can eliminate time by expressing, the velacity on the left side of the equation as v- dx/dt Then, we arrive at the following expression By integrating and applying the limits, we arrive at the following -vo which, sows that velacity decreases in a linear maner.
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