A particle A moves in one direction along a given trajectory with a tangential acceleration `w_tau=atau`, where a is a constant vector coinciding in direction with the x axis(figure), and `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

`v=sqrt(2ax)`

Explanation

In accordance with the problem
`w_t=veca*vectau`
But `w_t=(vdv)/(ds)` or `vdv=w_tds`
So, `vdv=(veca*vectau)ds=veca*dvecr`
or, `vdv=aveci*dvecr=adx` (because `veca` is directed towards the x-axis)
So, `underset0oversetvintdv=aunderset(0)overset(x)intdx`
Hence `v^2=2ax` or, `v=sqrt(2ax)`