A skydiver and her parachute weight 200 pounds. At the instant the parachute ope

Question

A skydiver and her parachute weight 200 pounds. At the instant the parachute opens, she is traveling at 28 feet per second. Assume that the air resistance varies directly as the instantaneous velocity and that the air resistance is 1000 pounds when the celerity is 20 feet per second.
(a) find the position of the skydiver at any time t after the parachute opens
(b) find the velocity of the skydiver at nay time t after the parachute opens
(c) find the limiting velocity

Please show me how to set this up. I know that the correct answers are:
(a) x = 4t-3e^(-8t)
(b) x = 4t+24e^(-8t)
(c) 4

Thanks so much!

Explanation / Answer

weight w = 200 lb, mass m = w/g.

downward = +x

t = 0, v0 = 28 ft/s, x(0) = 0,

the air resistance F = -kv, when v = 20 ft/s, F = 1000 lb, so k = 1000/20 = 50 lb/ft

mg - kv = mdv/dt

dv/(g - bv) = dt, where b = k/m = 50/(200/32) = 8

integrate, ln|g - bv| = -bt + C,

ln|32 - 8v| = -8t + C

C = ln|g - bv0| = ln|32 - 8*28| = ln(192)

ln|32 - 8v| = ln(192) - 8t

8v - 32 = 192 e-8t

v(t) = 4 + 28 e-8t (the velocity of the skydiver at nay time t after the parachute opens)

dx/dt = 4 + 28 e-8t   

integrate, x(t) = 4t - 3 e-8t (position of the skydiver at any time t after the parachute opens)

at the limiting velocity, dv/dt = 0, mg = kv, v = mg/k = g/b = 32/8 = 4 ft/s

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