Answer:
a) v(t) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K
where K = k/m
v(t) = [((kv₀/m) - g) e⁻ᴷᵗ + g] (m/k)
b) Terminal velocity = mg/k
c) s(t) = (gt/K) - [(Kv₀ - g) e⁻ᴷᵗ]
where K = k/m
s(t) = (mgt/k) - [((kv₀/m) - g))e⁻ᴷᵗ]
Step-by-step explanation:
m(dv/dt) = mg-kv
Divide through by m
(dv/dt) = g - (k/m)v
Let k/m = K
dv/dt = g-Kv
dv/(g - Kv) = dt
∫ dv/(g - Kv) = ∫ dt
∫ dv/(Kv - g) = - ∫ dt
(1/K) In (Kv - g) = -t + c (where c = constant of integration)
In (Kv - g) = - Kt + Kc
At t = 0, v = v₀
In (Kv₀ - g) = Kc
c = (1/K) In (Kv₀ - g)
In (Kv - g) = - Kt + Kc becomes
In (Kv - g) = - Kt + In (Kv₀ - g)
In (Kv - g) = [In (Kv₀ - g)] - Kt
Kv - g = (Kv₀ - g) e⁻ᴷᵗ
Kv = (Kv₀ - g) e⁻ᴷᵗ + g
v(t) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K
K = k/m
v(t) = [((kv₀/m) - g) e⁻ᴷᵗ + g] (m/k)
b) At terminal velocity, dv/dt = 0
From the starting differential equation,
m(dv/dt) = mg-kv
mg - kv = 0
kv = mg
v = mg/k
c) v = ds/dt = [(Kv₀ - g) e⁻ᴷᵗ + g]/K
(ds/dt) = [(Kv₀ - g) e⁻ᴷᵗ + g]/K
Kds = [(Kv₀ - g) e⁻ᴷᵗ + g] dt
∫ Kds = ∫ [(Kv₀ - g) e⁻ᴷᵗ + g] dt
Ks = -K((Kv₀ - g) e⁻ᴷᵗ) + gt
s(t) = (gt/K) - [(Kv₀ - g) e⁻ᴷᵗ]
K = k/m
s = (mgt/k) - [((kv₀/m) - g))e⁻ᴷᵗ]
