Two particles `A` and `B` are moving in the `x-y` plane such that their velocity components are `v_(x) = 1 m//s, v_(y) = (1)/(sqrt(3)) m//s` (for `A`) and `v_(x) = 2m//s, v_(y) = 2 m//s` (for `B`). If both the particles start moving from the same point, what is the angle between their paths ?

Explanation

`tan alpha = (1)/(1//sqrt(3)) = sqrt(3) rArr alpha = 60^(@)`
`tan beta = (2)/(2) = 1 rArr beta = 45^(@)`
Particles are moving in straight lines, hence the angle between their paths `= alpha - beta = 60^(@) - 45^(@) = 15^(@)`.