U-substitution for finding v(t)

[sadakaa]

Homework Statement

A .25kg ball is dropped from rest. Air resistance is f= -.05v
a. find v(t)
b. What is the terminal velocity of the ball?
c. At what time does the ball reach 50% of its terminal velocity? 90%?

Homework Equations

F=ma
F= mg - .05v

The Attempt at a Solution

Began to solve the equation knowing that I have to separate the terms, moves all the velocity terms to one side, then integrate.

ma = mg - .05v
a = mg/m - .05v/m
a = g - .2v

dv/dt = g - .2v
dt = dv/(g-.2v)

now I am just wondering to integrate that terms on the right. U-substitution for the denominator?

 

[PiratePhysicist]

\int{\frac{du}{a+k u}}=\frac{ln(a+k u)}{k}
Integral tables are your friend ^_^

 

[sadakaa]

so just to make sure I am correct, \int\frac{dv}{g-.2v} is \frac{ln(g-.2v)}{.2} + constant

 

[sadakaa]

alright so i assumed that was correct and went on with the problem, but ran into trouble at c.

ΣF = ma = mg - .05v
a = mg/m - .05v/.25
a = g - .2v
dv/dt = a

dv/dt = g - .2v
dv/(g - .2v) = dt
∫ dv/(g - .2v) = ∫dt
ln(g - .2v)/.2 = t - c
ln(g - .2v) = .2(t - c)
e^{.2(t - c)} = g - .2v
e^{.2(t - c)} - g = -.2v
-[e^{.2(t - c)} - g]/.2 = v(t)

Ball starts at rest, v = 0 when t = 0
[e^{-.2c} - g]/-.2 = 0
-5[e^{-.2c} - g] = 0
-5e^{-.2c} + 5g = 0
-5e^{-.2c} = -5g
e^{-.2c} = g
ln(g) = -.2c
-ln(g)/.2 = c
c = -11.42

v(t) = -5e.2(t + 11.42) + 49.05

What is the terminal velocity?
ΣF = ma = mg - .05v = 0
.05v = .25g
v = 5g = 49.05 m/s

At what time does the ball reach 50% of its terminal velocity?
v(t) = -5e.2(t + 11.42) + 49.05
v(t) = 24.525 m/s

24.525 - 49.05 = -5e^{.2(t + 11.42)}
4.905 = e^{.2(t + 11.42)}
ln(4.905) = .2(t + 11.42)
ln(4.905) - 2.284 = .2t
[ln(4.905) - 2.284]/.2 = t

t = -3.47 sec
t is negative. whattttt?