Question

In (14) of Section 1.3 we saw that a differential equation describing the velocity v of a falling mass subject to air resistance proportional to the instantaneous velocity is m dv/dt = mg - kv, where
k > 0 is a constant of proportionality. The positive direction is downward.

a. Solve the equation subject to the initial condition v(0) = v0.
vt = _____

b. Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass.

c. If the distance s, measured from the point where the mass was released above ground, is related to velocity v by ds/dt = v(t),find an explicit expression for s(t) if s(0) = 0.

Answer 1

a)

m$\frac {dv}{dt}$ = mg - kv where k>0 is a constant

$m\frac {dv}{dt}$ = - ( kv - mg)

$\frac {dv}{dt}$ = - $\frac {k}{m}$ (v - $\frac {mg}{k}$ )

=> $\frac {dv}{v - \frac{mg}{k}}$ = - $\frac {k}{m}$ dt

Integrating

$\int$ $\frac {dv}{v - \frac{mg}{k}}$ = -$\frac {k}{m}$ $\int$ dt

=> ln(v-$\frac {mg}{k}$ ) = -$\frac {k}{m}$ t + C, C is integration constant

By initial condition, when t = 0, v = v₀

So C = ln(v₀ - $\frac {mg}{k}$ )

Therefore

ln(v - $\frac {mg}{k}$ ) = - $\frac {k}{m}$ t + ln(v₀ - $\frac {mg}{k}$ )

=> ln $(\frac { v - \frac {mg}{k}}{v_0 - \frac {mg}{k}})$ = - $\frac {k}{m}$ t

=> $(\frac { v - \frac {mg}{k}}{v_0 - \frac {mg}{k}})$ = $e^{-\frac{k}{m}t}$

=> $v - \frac {mg}{k}$ = $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$

=> $v$ = $\frac {mg}{k}$ + $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$

So $v(t) =$ $\frac {mg}{k}$ + $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$

b)

For terminal velocity t→∞

So terminal velocity =

$\lim_{t→∞}v(t)$

= $\lim_{t→∞}$ [$\frac {mg}{k}$ + $(v_0 - \frac {mg}{k})$$e^{-\frac{k}{m}t}$ ]

= $\frac {mg}{k}$ because $e^{-\frac{k}{m}t}$ →0 as t→∞

So terminal velocity is $\frac {mg}{k}$

c)

$\frac {ds}{dt}$ = $\frac {mg}{k}$ + $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$

$\int$ ds = $\int$ [$\frac {mg}{k}$ + $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$ ]dt

=> s = $\frac {mg}{k}t$ - $\frac{m}{k}$ $(v_0 - \frac {mg}{k})$ $e^{-\frac{k}{m}t}$ + D,

D is integration constant

From initial condition, s = 0, when t = 0

So D = $\frac{m}{k}$ $(v_0 - \frac {mg}{k})$

s = $\frac {mg}{k}t$ + $\frac {m}{k}$ $(v_0 - \frac {mg}{k})(1-e^{-\frac{k}{m}t})$

s(t) =

$\frac {mg}{k}t$ + $\frac {m}{k}$ (v₀- $\frac {mg}{k}$ )(1- $e^{-\frac{k}{m}t}$ )