Question

Imagine that an object of mass m is initially moving in the x direction through a viscous fluid at a speed v. If the only force acting on this object is a viscous drag force whose magnitude is given by F=bv, prove that the object's x-velocity is given by vx(t)=ve^-qt and determine how the constant q depends on b and m.

Answer 1

Hint: This is actually a motion-from-force problem. Divide both sides of the x component of Newton's second law by v, and take the indefinite integral of both sides

Answer 2

alright well lets do the hint then

newtons second law
F = ma

here F is bv

bv = ma

which is

b/m x dx/dt = d^2x/dt^2

dx/dt = vx(t)=ve^-qt
so d^2x/dt^2 = -vqe^-qt

plugging this back in

b/m x ve^-qt = -vqe^-qt

so we can see for this to hold -q = b/m

not sure how to do it integrating.

Answer 3

F=bv
which acts opposite to F= m a (Newton's second law)
write "a" as dv/dt
\[ b v = -m \frac{dv}{dt } \\ \frac{b}{m} v = \frac{dv}{dt } \]
for ease of typing, let k= b/m
also, "split" the infinitesimal:
\[ -k \ dt = \frac{dv}{v} \]
integrate both sides
\[ \int \frac{dv}{v} = \int -k \ dt \\
\ln(v) = -kt + c_1\\
v = e^{-kt+c_1} \\
v= c_2 \ e^{-kt}
\]
where \( c_2 \) is the constant of integration \( e^{c_1} \)
at t=0 v= \( c_2 \) , and we see the constant is the initial velocity
denote the velocity at t=0 as \( v_0\), and replace -k with -b/m to
get the result
\[ v_x(t)= v_0 \ e^{-\frac{b}{m}t} \]