Airplanes `A` and `B` are flying with constant velocity in the same vertical plane at angles `30^(@)` and `60^(@)` with respect to the horizontal respectively as shown in figure . The speed of `A` is `100sqrt(3) m//s` . At time `t = 0 s`, an observer in `A` finds `B` at a distance of `500 m`. The observer sees `B` moving with a constant velocity perpendicular to the line of motion of `A` . If at `t = t_(0)`, A just escapes being hit by `B ,t_(0)` , A just escapes being hit by `B , t_(0)` in seconds is

Answer

5

Explanation

[5]: See the following figure for velocitys of planes `(vecv_(a) and vecv_(B))` and relative velocity of B with respect to A `(vecv _("BA"))`.

`vecv_(A) =(100 sqrt3cos 30) hati+(100 sqrt3 sin 30)hatj=150hati+50 sqrt3hatj`
`vecv_(B)=(v_(B) cos 60)hati+(v_(B) sin 60)hatj=(v_(B))/(2)hati+(v_(B)sqrt3)/(2)hatj`
Velocity of B with respect to A can be written as follows:
`vecv_("BA")=vecv_(B)-vec(v)_(A)`
`rArr vecv_(BA)=((v_(B))/(2)-150) hati+sqrt3((v_(B))/(2)-50)hatj`
It is given tyhat velocity of B with respect to A is perpendicul,ar to A, hence dot product of these velocities must be zero .
`vecv_(BA). vec(c)_(A) =0`
`rArr 150 ((v_(B))/(2)-150)+150((v_(B))/(2)-50)=0`
`rArr v_(B)=200m//s`
By substituting value of `v_(B)` in equation (i) , we can write velocity of B with respect to A as follows:
`rArr vecv_(BA)=(-50hati+50sqrt3hatj)m//s`
`rArr v_(BA)=100m//s`
Initial distance between the two is given to be 500 m, hence time of collision can be written as
`t_(0)=(500)/(100)=5s`
Hence answer is 5 .