Step-by-step explanation:
a) The general solution is given to be:
v = -mg/c + Ae^(-ct/m)
To derive this, start with the differential equation:
m dv/dt = -mg − cv
Separate the variables:
dv / (mg + cv) = -1/m dt
c dv / (mg + cv) = -c/m dt
Integrate:
ln(mg + cv) = -c/m t + A
mg + cv = e^(-c/m t + A)
mg + cv = Ae^(-c/m t)
cv = -mg + Ae^(c/m t)
v = -mg/c + Ae^(c/m t)
*Notice that A is an arbitrary constant, so multiplying it by another constant is still an arbitrary constant.
At t=0, v=0:
0 = -mg/c + A
A = mg/c
Substitute:
v = -mg/c + mg/c e^(-ct/m)
v = -mg/c (1 − e^(-ct/m))
Given that mg = 176 and m = 11/2:
v = -176/c (1 − e^(-2ct/11))
b) The terminal velocity occurs as t approaches infinity.
-120 = -176/c (1 − 0)
c = 176/120
c = 22/15
c) Plugging in the value for c, the velocity equation is:
v = -120 (1 − e^(-4t/15))
At t = 10, the velocity is:
v = -120 (1 − e^(-40/15))
v = -111.66 ft/s
d) To find the distance, integrate from t=0 to t=10.
d = ∫₀¹⁰ v dt
d = ∫₀¹⁰ -120 (1 − e^(-4t/15)) dt
d = ∫₀¹⁰ (-120 + 120 e^(-4t/15)) dt
d = ∫₀¹⁰ (-120 − 450 × -4/15 e^(-4t/15)) dt
d = (-120t − 450 e^(-4t/15)) |₀¹⁰
d = (-1200 − 450 e^(-40/15)) − (0 − 450)
d = -750 − 450 e^(-8/3)
d = -781.27 ft