Two particles are moving along two long straight lines, in the same plane with same speed equal to `20 cm//s.` The angle between the two linse is `60^@` and their intersection point isO. At a certain moment, the two particles are located at distances 3m and 4m from O and are moving twowards O. Subsequently, the shortest distance between them will be

Answer

`50sqrt(3)cm`

Explanation

`vecV_(AB) = vecV_(A) - vecV_(B)`
`rArr V_(AB) = sqrt(V_(A)^(2) +V_(B)^(2) +2V_(A) (-V_(B))cos 120^(@))`
`rArr V(AB) = V_(A)`
for shortest distance we draw perpendicular on the line of motion of `A` relative to Bfrom `B`.
`sin 60^(@) = (x)/(1) rArr`