The differential equation governing the velocity v of a falling mass m subject to air resistance...?

Question

proportional to the instantaneous velocity is:

m*dv/dt = mg - kv

where k is a positive constant of proportionality.

a) Solve the equation subject to the initial condition v(0)=A (a constant)

b) Determine the limiting, or terminal, velocity of the weight.

c) if the distance, s, is related to the velocity ds/dt = v, find an explicit expression s if it is further known that s(0)=B (a constant)

Can anyone help? I've been stuck on this problem for a while...

kb · Accepted Answer

a) Rewrite in standard form (for a first order linear DE):
m dv/dt + kv = mg
==> dv/dt + (k/m)v = g

Multiply both sides by the integrating factor e^(kt/m):
e^(kt/m) dv/dt + (k/m) e^(kt/m)v = ge^(kt/m)
==> (d/dt) [e^(kt/m) v] = ge^(kt/m), by the product rule.

Integrate both sides:
e^(kt/m) v = (mg/k) e^(kt/m) + C
==> v(t) = mg/k + Ce^(-kt/m).

Find C via v(0) = A:
A = mg/k + C * 1 ==> C = A - mg/k.

Hence, v(t) = mg/k + (A - mg/k) e^(-kt/m).
----------------
b) Let t→∞ with the result from (a):
lim(t→∞) [mg/k + (A - mg/k) e^(-kt/m)] = mg/k + (A - mg/k) * 0 = mg/k.
----------------
c) Integrate the result from (a):
s(t) = mgt/k + (A - mg/k) * (-m/k)e^(-kt/m) + C

We find C by using s(0) = B:
B = 0 + (A - mg/k) * (-m/k) * 1 + C
==> C = B + (A - mg/k) * (m/k).

Hence, s(t) = mgt/k + (A - mg/k) * (-m/k)e^(-kt/m) + [B + (A - mg/k) * (m/k)].

I hope this helps!

? · Answer

some hints for you

m*dv/dt = mg - kv
dv/dt = g - kv / m
take the anti - derivative with respect to v
v = gt - kvt / m + A 
where A is a constant

v + kvt / m = gt + A 
v(1 + kt / m) = gt + A
v = (gt + A) / (1 + kt / m)

s = Integral[(gt + A) / (1 + kt / m) dt]

s = m((Ak - gm)ln(kt + m) + gkt) / k^2 + constant

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The differential equation governing the velocity v of a falling mass m subject to air resistance...?

2 Answers