A man wants to reach from A to the opposite corner of the square C. The sides of the square are 100 m. A central square of `50mxx50m` is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square , he can walk only at a speed of v m/s `(v lt 1)`. What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand ?

Options

A. 0.18 m `s^(-1)`
B. 0.81 m `s^(-1)`
C. 0.5 m `s^(-1)`
D. 0.95 m `s^(-1)`

Answer

B

Explanation

`AC=sqrt((100)^(2)+(100)^(2))=100sqrt(2)m`
`PQ=sqrt((50)^(2)+(50)^(2))=50sqrt(2)m`
`AP=(AC-PQ)/(2)=(100sqrt(2)-50sqrt(2))/(2)`
`=25sqrt2m`
`QC=AP=25sqrt(2)m,AR=sqrt((75)^(2)+(25)^(2))=25sqrt(10)m`
`RC=AR=25sqrt(10)m`
Consider the straight line APQC through the sand. Time taken to go from A to C via this path
`T_("sand")=(AP+QC)/(1)+(PQ)/(v)`
`=(25sqrt(2)+25sqrt(2))/(1)+(50sqrt(2))/(v)=50sqrt(2)[(1)/(v)+1]`
The shortes path outside the sand will be ARC. Time taken to go from A to C via this path
`T_("outside")=(AR+RC)/(1)=(25sqrt(10)+25sqrt(10))/(1)=50sqrt(10)`
`:'T_("sand")ltT_("outside"):.50sqrt(2)[(1)/(v)+1]lt50sqrt(10)`
`implies(1)/(v)+1ltsqrt(5)or(1)/(v)ltsqrt(5)-1orvgt(1)/(sqrt(5)-1)~~0.81ms^(-1)`