Two body A and B are 10km apart such that B is in south of A. A and B start moving with same speed 20km/hr eastward respectively then find.
(a) Relative velocity of A w.r.t. B. (b) Minimum separation atained during motion
(c) Time lapse, from starting to attain minimum separation.

Explanation

In the adjoining figure are shown the total acceleration vector `vec(a)` and its components the tangential accelerations `vec(a)_(tau)` and normal acceleration `vec(a)_(n)` are shown. These two components are always mutually perpendicular to each other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector makes an angle of `45^(@)` with the radius, both the tangential and the normal components must be equal in magnitudr.
Now from equation and, we have
`a_(tau)=a_(n) rarr alpha R=omega^(2)R rArr alpha=omega^(2)` ...(i)
Since angular acceleration is uniform, form equation, we have `omega=omega_(o)+alphat`
Substituting `omega_(0)=0` and `t=2 s`, we have `omega=2alpha` ...(ii)
From equation (i) and (ii), we have `alpha=0.25 rad//s^(2)`