What is happening in this step, kinematics problem

[wahaj]

Homework Statement

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

Homework Equations

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

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.

 

[Mark44]

Homework Statement

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

Homework Equations

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

Typo above - it should be dx in the first integral.

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.

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).

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.

 

[wahaj]

did your post cut off or something?

 

[Dick]

did your post cut off or something?

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.

 

[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.

 

[Dick]

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.

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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

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.

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.