OB and OC are two roads enclosing an angle of `120^(@)`. X and Y start along OB with a speed of `4kmh^(-1)` and Y travels alog OC with a speed of `3kmh^(-1)`. The rate at which the shortest distance between X and Y is increasing after 1h is

Options

A. `sqrt(37)kmh^(-1)`
B. `37kmh^(-1)`
C. `13kmh^(-1)`
D. `sqrt(13)kmh^(-1)`

Answer

A

Explanation

After time `t` the distance covered by `X` is `4t` and `Y `is `3t`.
Let shortest distance between X and Y is A. Then, by cosine law
`A^(2)=(4t)^(2)+(3t)^(2)-(4t)(3t)2 cos 120^(@)`
`impliesA^(2)=16t^(2)-9t^(2)-24t^(2)(-1/2)=37t^(2)`…………….i
`impliesA=sqrt(37t)`
If `t=1h,` then `A=sqrt(37)km`
Now, differentiating Eq. (i) w.r.t `t` we get
`2"AA"'=37(2t)`
After `t=1h`, we get
`2sqrt(37)A'=2(37)impliesA'=sqrt(37)km//h`