Consider the motion of a bacterium \"B\" moving in a viscous fluid. f B is the a

Question

Consider the motion of a bacterium "B" moving in a viscous fluid. f B is the amount of force B generates to continue swimming. If B stops providing the force fB at time-0: 1) Write a general expression for distance x(t) that B will coast, given initial speed of Vo 2) If B's 00 30 s , (microns per second), and b3 3) If B is 1 m long, what is the distance B coasts in units of its body size? And write V0 4) Repeat parts 1, 2, and 3 for person C coasting in the same fluid, with C weighing about = [ ,(microseconds), find the total distance B coasts in units of its body size per second. 10 14 as much as B, swimming at v0 -1 m/s. Since C is much larger, the equation for swimming force is: C mx+b(x2 and b 0.1 m. Also find the distance C coasts in the time it takes its speed to reduce by 1% of the initial to . c is about 1 m tall

Explanation / Answer

given fb = 0 for t > 0

then

mx" + bx' = 0

intiial speed = vo

mdv/dt + bv = 0

mdv/dx dx/dt + bv = 0

mvdv = -bvdx

mdv = -bdx

integrating

m(v - vo) = -b(x)

mdv/dt = -bv

a. mv = mvo - bx = b(mvo/b - x)

mdx/(x - mvo/b) = -bdt

integrating

ln(xb/mvo - 1) = -bt/m

x = mvo(exp(-bt/m) + 1)/b

b. vo = 30 um/s

m/b = 1/3 us

then as t -> inf,

x = mvo/b = (1/3)(30) = 10 um*10^-9 = 10^-8 um

c. l = 1um

vo = 30 body units per sec

x = 10^-8 body units

d. weight is increased to 10^14 timnes

vo = 1 m/s

f = mx" + bx^2

f = 0

mdv/dt = -bx^2

mdv/dt dx/dx = m v dv = -bx^2*dx

integrating

m(v^2 - vo^2)/2 = -b(x^3)/2

solving this equation we can find dependence of v on t and x on t

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