v(t) with air resistance

KindofSlow

Junior Member
Joined
Mar 5, 2010
Messages
90
Problem: Air resistance on falling body is a = dv/dt = g-kv with k is constant. Find v(t) assuming starts from rest.
Here's my work.
dv/dt = g-kv
dv = (g-kv)dt
1/(g-kv)dv = dt
∫ 1/(g-kv)dv = ∫dt
u = (g-kv) , du/dv = -k , -1/k du = dv
-1/k ∫ 1/u du = ∫ dt
(-1/k)(ln(g-kv)) = t
ln(g-kv) = -kt
g-kv = e^(-kt)
kv = g - e^(-kt)
v = 1/k(g-e^(-kt))
Correct answer according to book is v = g/k(1-e^(-kt))

I feel like I've gotten close but I must have a mistake somewhere and I cannot find it.
If anyone can point out my error, I would appreciated it.
Thank you
 
Problem: Air resistance on falling body is a = dv/dt = g-kv with k is constant. Find v(t) assuming starts from rest.
Here's my work.
dv/dt = g-kv
dv = (g-kv)dt
1/(g-kv)dv = dt
∫ 1/(g-kv)dv = ∫dt
u = (g-kv) , du/dv = -k , -1/k du = dv
-1/k ∫ 1/u du = ∫ dt
(-1/k)(ln(g-kv)) = t + C1

You are forgetting constant of integration.

ln(g-kv) = -kt
g-kv = e^(-kt)
kv = g - e^(-kt)
v = 1/k(g-e^(-kt))
Correct answer according to book is v = g/k(1-e^(-kt))

I feel like I've gotten close but I must have a mistake somewhere and I cannot find it.
If anyone can point out my error, I would appreciated it.
Thank you

.
 
Subhotosh, I apologize, now I don't know how to properly deal with the constant of integration.
My attempt:
(-1/k)(ln(g-kv)) = t + c
ln(g-kv) = -kt + c ; (-k*c = c#2)
g-kv = c*e^(-kt) ; (e^c = c#3)
kv = g - ce^(-kt)
v = g/k - (c)e^(-kt) ; (c/k = c#4)
I'm on my 4th constant and just don't know how to deal with this.
Thank you for your help.

 
Subhotosh, I apologize, now I don't know how to properly deal with the constant of integration.
My attempt:
(-1/k)(ln(g-kv)) = t + c
ln(g-kv) = -kt + c ; (-k*c = c#2)
g-kv = c*e^(-kt) ; (e^c = c#3)
kv = g - ce^(-kt)
v = g/k - (c)e^(-kt) ; (c/k = c#4)
I'm on my 4th constant and just don't know how to deal with this.
Thank you for your help.

Find v(t) assuming starts from rest

v = 0 at t=0 → 0 = g/k - c * ek*0 → c = g/k
 
Top