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Find the particle's horizontal position x(t) and velocity v(x) at any point 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. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F-F = m Then: -m v= m

Find the particle's horizontal position x(t) and velocity v(x) at any point 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. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F-F = m Then: -m v= m

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Find the particle's horizontal position x(t) and velocity v(x) at any point 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. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F=F = m Then: -m v = m since: a - dv/dt then -m v = m by integrating, we obtain the following expression: s voe Further, employing the rules of integration results to the following expression for position as a function of time x= (vo/ - e as t+ 0, the position becomes x= vo/k b) To solve for the velocity as a function of position v(x), construct the net force in the x-axis as follows EF-F sm Then: -m v = m since: a = dv/dt then -m V= m We can eliminate time by expressing, the velocity on the left side of the equation as V= dx/dt Then, we arrive at the following expression = -k By integrating and applying the limits, we arrive at the following = VO which, sows that velocity decreases in a linear maner.
Transcribed Image Text:Find the particle's horizontal position x(t) and velocity v(x) at any point 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. Consider that the particle is initially traveling with a velocity vo. Solution: a) To solve for the position as a function of time x(t), we construct the net force in the x-axis as E F=F = m Then: -m v = m since: a - dv/dt then -m v = m by integrating, we obtain the following expression: s voe Further, employing the rules of integration results to the following expression for position as a function of time x= (vo/ - e as t+ 0, the position becomes x= vo/k b) To solve for the velocity as a function of position v(x), construct the net force in the x-axis as follows EF-F sm Then: -m v = m since: a = dv/dt then -m V= m We can eliminate time by expressing, the velocity on the left side of the equation as V= dx/dt Then, we arrive at the following expression = -k By integrating and applying the limits, we arrive at the following = VO which, sows that velocity decreases in a linear maner.
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