This is exercise 11 of section 14.15 of Apostol Calculus:
A particle moves along a plane curve with constant speed $5$. It starts at the origin at time $t=0$ with initial velocity $5j$, and it never goes to the left of the $y$-axis. At every instant the curvature of the path is $k(t)=2t$. Let $\alpha(t)$ denote the angle that the velocity vector makes with the positive $x$-axis at time $t$. Determine $\alpha(t)$ explicitly as a function of $t$.
Because $k(t) = |\frac{d\alpha}{ds}|$ where $s$ is the arc length we can use $$\frac{d\alpha}{dt}=\frac{d\alpha}{ds}\cdot\frac{ds}{dt} = \frac{d\alpha}{ds}\|v(t)\|$$
to get the differential equation $$|\alpha'| =10t$$
Because we need not to go on the left of the $y$-axis and $\alpha(0)= \frac{\pi}{2}$ we need $\alpha' <0$ for a certain amount of time.
In the back of the book it's written the solution $\alpha(t) = \frac{\pi}{2} -5t^2$.
I think this solution requires $\alpha'$ less than zero for all $t$. However, it seems to me that it's possible that $\alpha$ increases after some time so that the angle could have a more complicated form.
Why is it instead impossible?