Question

A particle $A$ moves in one direction along a given trajectory with a tangential acceleration ${\omega }_{\tau }=a\tau$, where $\overrightarrow{a}$ is a constant vector coinciding in direction with the $x$ axis as shown in figure above, and $\overrightarrow{\tau }$ 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 $x=0$.

Answer 1

In accordance with the problem
$w_t=\overrightarrow{a}\cdot \overrightarrow{\tau }$
But $w_t=\frac{vdv}{ds}$ or $vdv=w_{t}ds$
So, $vdv=(\overrightarrow{a}\cdot \overrightarrow{\tau })ds=\overrightarrow{a}\cdot d\overrightarrow{r}$
or, $vdv=a\overrightarrow{i}\cdot d\overrightarrow{r}=a\cdot dx$ (because $\overrightarrow{a}$) is directed towards the x-axis)
So, ${\int }_0^{v}vdv=a{\int }_0^{x}dx$
Hence $v^2=2ax$ or, $v=\sqrt{2ax}$