Okay so I've been working a mechanics problem and it has boiled down to this.
I want to find $v(t)$ and I currently have that.
$$t+c_1=\frac{1}{2\sqrt{gk}}\ln{\frac{\sqrt{g/k}-v}{\sqrt{g/k}+v}}$$ where $c_1$ is a constant of integration.
My thoughts are so, at $t=0$ the particle has some velocity $v$ so we will end up with $c_1=...$ but I'm not sure how to determine $c_1$ I think I need to sub in $t=0$ or something but I don't know how to do it.
Then obviously I need to rearrange to find $v(t)$ and then I am done, but this is easier said than done considering this equation is quite ugly to be honest. Any help with me on this one?
Thanks.
Isolate the log function to one side as follows:- $$2\sqrt{gk}(t+c_1)=ln{\frac{\sqrt{g/k}-v}{\sqrt{g/k}+v}}$$ then take the exponential of both sides $$\frac{\sqrt{g/k}-v}{\sqrt{g/k}+v}=e^{2\sqrt{gk}(t+c_1)}$$ leading to $v(t)$ being $$v(t)=\frac{\sqrt{g/k}(1-e^{2\sqrt{gk}(t+c_1)})}{1+e^{2\sqrt{gk}(t+c_1)}}$$
Now, in order to find the value of $c_1$ at time $t=0$ and $v(0)=V$, you need to make $c_1$ the subject of the formula, leading to $$c_1=\frac{1}{2\sqrt{gk}}\left(ln{\frac{\sqrt{g/k}-v}{\sqrt{g/k}+v}}-2t\sqrt{gk}\right)$$ Set $t=0$ and $v(0)=V$, and you are done.