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

`v = (-384)/(5) (1 - (12)/(5) e^( -5t/12))`

Step-by-step 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

`(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(32 + 5v/12)/(c) = -5t/12`

Take exponential of both sides

`(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 - (384)/(5) e^( -5t/12))\\v = (-384)/(5) (1 - (12)/(5) e^( -5t/12))`

Answer 2

Step-by-step explanation:

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