Moving in the same direction along parallel tracks, $A$ and $B$ pass the point $O$ simultaneously with speeds of $20\text{ cm/s}$ and $10\text{ cm/s}$ respectively. From then on, the deceleration of $A$ is $\dfrac{v^3}{400}\text{cm/s}^2$ while the deceleration of $B$ is $\dfrac{v^2}{100}\text{cm/s}^2$ when their speeds are $v\text{ cm/s}$. Find the speeds of $A$ and $B$ at any time $t$ seconds after passing $O$.
Show that $B$ overhauls $A$ during the $15^{th}$ second after passing $O$, but falls behind again during the $44^{th}$ second.
Here is my attempt ...
Deceleration of A
$$a_{A} = \frac{dv}{dt} = -\frac{v^3}{400} $$
$$\int_{20}^{v}\frac{dv}{v^3} = \int_{0}^{t} -\frac{dt}{400}$$
$$\left[-\frac{1}{2v^2}\right]_{20}^v = \left[-\frac{t}{400}\right]_0^t$$
$$\frac{1}{2v^2} - \frac{1}{800} = \frac{t}{400}$$
$$\frac{1}{v^2} = \frac{1}{400} + \frac{t}{200}$$
$$\frac{1}{v^2} = \frac{2 + 4t}{800}$$
$$v^2 = \frac{400}{1 + 2t}$$
$$v_A = \pm\sqrt{\frac{400}{1 + 2t}}$$
Deceleration of B
$$a_{B} = \frac{dv}{dt} = -\frac{v^2}{100} $$
$$\int_{10}^{v}\frac{dv}{v^2} = \int_{0}^{t} -\frac{dt}{100}$$
$$\left[-\frac{1}{v}\right]_{10}^v = \left[-\frac{t}{100}\right]_0^t$$
$$\frac{1}{v} - \frac{1}{10} = \frac{t}{100}$$
$$v_B = \frac{100}{t+10}$$
I hope I got the first part right. I guess, integrating V again would get me to the position. But then I guess I will end up in
$\log|\text{some expression}|$
Is this correct? Any help in solving the second part is greatly appreciated. Thank you.