A particle $$A$$ moves in one direction along a given trajectory with a tangential acceleration $$\omega_\tau=a\tau$$, where $$\vec{a}$$ is a constant vector coinciding in direction with the $$x$$ axis as shown in figure above, and $$\vec{\tau}$$ is a unit vector coinciding in direction with the velocity vector at a given point. Find how the velocity of the particle depends on $$x$$ provided that its velocity is negligible at the point $$x=0$$.

Solution

In accordance with the problem
$$w_t=\vec{a}\cdot\vec{\tau}$$
But $$\displaystyle w_t=\frac{vdv}{ds}$$ or $$vdv=w_tds$$
So, $$vdv=(\vec{a}\cdot\vec{\tau})ds=\vec{a}\cdot d\vec{r}$$
or, $$vdv=a\vec{i}\cdot d\vec{r}=a\cdot dx$$ (because $$\vec{a}$$) is directed towards the x-axis)
So, $$\displaystyle\int_0^v{vdv}=a\int_0^x{dx}$$
Hence $$v^2=2ax$$ or, $$v=\sqrt{2ax}$$