What is happening in this step, kinematics problem

1. Jan 15, 2013

wahaj

1. The problem statement, all variables and given/known data
a particle travels along a straight line and the acceleration is given by
a = 30 - 0.2v
determine the time when the velocity of the particle is v = 30

2. Relevant equations

$$\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b)$$

3. The attempt at a solution
This is the solution from the textbook. The problem is that my math class is behind schedule so I haven't learned integrals yet. The only knowledge I have on the matter is from what was briefly covered in my dynamics class. So I need an explanation of what is happening is happening in this solution.
$$dt = \frac{dv}{a}$$
$$\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v}$$
$$t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0$$
I understand everything upto this point but I have no idea what is happening in the next step.
$$t = 5 \ln \frac {30}{30-0.2v}$$
$$t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s$$
several questions, first where did the 30 in the numerator come from? why did the minus sign before $$\frac{-1}{0.2} = -5$$ go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.

2. Jan 15, 2013

Staff: Mentor

Typo above - it should be dx in the first integral.
The right side above is -5[ln(30 - .2v) - ln(30)] = 5[ln(30) - ln(30 - .2v)]
$$= 5 ln\frac{30}{30 - .2v}$$
They're just using a property of logs; namely that ln(a/b) = ln(a) - ln(b).

3. Jan 15, 2013

wahaj

did your post cut off or something?

4. Jan 15, 2013

Dick

No, I think it's all there. Look at the part before the last quote. That's answering the question of why 5 instead of -5 and the manipulation of the logs stuff.

5. Jan 15, 2013

wahaj

well i managed to figure it out when I found out that the log property was used. I don't understand the point of putting the variable in the denominator instead of simply writing
$$\ln \frac{30-0.2v}{30}$$
why make things unnecessarily complicated just to get rid of a negative sign.
Anyways thanks for the help.

6. Jan 15, 2013

Dick

Yes, there was no NEED to get rid of the negative. They just chose to write it that way.

7. Jan 16, 2013

Ray Vickson

There is some vital information missing from the problem statement: what is the initial velocity v0 at t = 0? For example, if v0 = 50 we cannot have v(t) = 30 for t > 0; the solution has t < 0; if v0 > 150 there is no time where v(t) = 30.