Question

A firefighter who weighs 192 lb slides down an infinitely long fire pole that exerts a friction resistive force with magnitude proportional to his speed, with k D 2:5 lb-s/ft. Assuming that he starts from rest, find his velocity as a function of time and find his terminal velocity

Answer 1

Answer:

Explanation:

check the pictures attached to further understand and i hope it works

Answer 2

Answer:

$v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})$

Explanation:

Weight of the firefighter, W = 192 lb

W = mg

g = 32 ft/s²

Mass of the firefighter, m = W/g

m = 192/32

m = 6 slugs

k = 2.5 lb-s/ft

The force, F = ma = kv

2.5v = 6a

a = 2.5v/6

a = 5v/12

The fundamental dynamic equation;

dv/dt + drag + gravity = 0

dv/dt = -g-a

dv/dt = -32-5v/12..............(a)

The motion will attain terminal velocity when dv/dt = 0

-32 - 5v/12 = 0

-32 = 5v/12

-384 = 5v

v₀ = -384/5

v₀ = -384/5

dv/dt = -32 - 5v/12

$\frac{dv}{-32 -5v/12} = dt$

$-12/5 ln(32 + 5v/12) = t + c\\ln(32 + 5v/12) = -5t/12 + lnc\\ln(32 + 5v/12) - ln c = -5t/12\\ln\frac{32 + 5v/12}{c} = -5t/12$

Take exponential of both sides

$\frac{32 + 5v/12}{c} =e^{ -5t/12}\\32 + 5v/12 = ce^{ -5t/12}\\5v/12 = -32 + ce^{ -5t/12}\\v = 12/5 (-32 + ce^{ -5t/12})$

c = v₀ = -384/5

$v = 12/5 (-32 - \frac{384}{5} e^{ -5t/12})\\v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})$