# B Integrating velocity equation problem

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1. Apr 10, 2016

### Sho Kano

I've already completed the first question, but with number two, it's a different case. Here's my attempt:
$\frac { d{ v }_{ y } }{ dt } \quad =\quad -g\quad -\quad \beta { v }_{ y }\\ \frac { d{ v }_{ y } }{ -g\quad -\quad \beta { v }_{ y } } \quad =\quad dt\\ \int { \frac { d{ v }_{ y } }{ -g\quad -\quad \beta { v }_{ y } } } \quad =\quad \int { dt } \\ \frac { -1 }{ \beta } ln(-g-\beta { v }_{ y })\quad =\quad t\\$
natural log of a negative?

2. Apr 10, 2016

### Sho Kano

EDIT: I seem to have figured out the answer right after posting this?
$\frac { -1 }{ \beta } ln(-g-\beta { v }_{ y })\quad =\quad t\\ ln(-g-\beta { v }_{ y })\quad =\quad -t\beta \\ -g-\beta { v }_{ y }\quad =\quad { e }^{ -t\beta }$

3. Apr 10, 2016

### Ssnow

In general $\int \frac{1}{x} dx= \log{|x|}+c$

4. Apr 10, 2016

### Sho Kano

log base e right?

5. Apr 10, 2016

### Ssnow

yes, $\int \frac{1}{x}=\ln{|x|}+c$

6. Apr 10, 2016

### Ssnow

if $-g-\beta v_{y}>0$ this is part of your solution, you must use the condition that $v_{y0}=0$, but attention that you forgot the integration constant in your solution ...

7. Apr 10, 2016

### Sho Kano

The integration constant is just the initial condition, which is 0

8. Apr 10, 2016

### Ssnow

mmm, I don't think, assuming $-g-\beta v_{y}>0$ put the condition $v_{y}=0$ with $t=0$ in

$\ln{(-g-\beta v_{y})}=-t\beta +c$

9. Apr 10, 2016

### Ssnow

You can see the value of $c$ also from $\ln{|-g-\beta v_{y}|}=-t\beta + c$...

10. Apr 10, 2016

### Sho Kano

Ssnow, not sure I follow you. From post #1, 3rd line of my attempt, the limits of integration are:

left side: 0 to v
(its 0 since initial velocity is given to be 0)
right side: 0 to t

11. Apr 11, 2016

### Ssnow

You forget the constant $\ln{g}$:

$\int_{0}^{v_{y}}\frac{w}{-g-\beta w}=-\frac{1}{\beta}\ln{|-g-\beta v_{y}|}+\frac{1}{\beta}\ln{g}$

12. Apr 11, 2016

### Sho Kano

I see now, thanks!