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 ?

Explanation

Consider adjacent diagram

Time taken to go from A to C via straight line path APQC through the S and
`T_("sand")=(AP+QC)/(1)+(PQ)/(v)=(25sqrt(2)+25 sqrt(2))/(1)+(50 sqrt(2))/(v)`
`=50 sqrt(2)+ 50sqrt(2)/(v)=50 sqrt(2)((1)/(v)+1)`
Clearly from figure the shortest path outside the sand will be ARC.
Time taken to go from A to C via this path
`T_("outside")=(AR+RC)/(1)s`
`"Clearly "AR=sqrt(75^(2)+25^(2))=sqrt(75xx75+25xx25)`
`=5xx5sqrt(9+1)=25sqrt(10)m`
`RC=AR=sqrt(75^(2)+25^(2))=25sqrt(10)m`
`rArr " " T_("outside")=2AR=2xx25 sqrt(10)s=50 sqrt(10)s`
`"For "T_("sand")lt T_("outside")`
`rArr " " 50sqrt(2)((1)/(v)+1)lt 2 xx 25 sqrt(10)`
`rArr " " (2sqrt(2))/(2)((1)/(v)+1)lt sqrt(10)`
`rArr " " (1)/(v)+1lt(2sqrt(10))/(2sqrt(2))=(sqrt(5))/(2)xx2=sqrt(5)`
`rArr " " (1)/(v)lt(sqrt(5))/(2)xx2-1rArr(1)/(v)lt sqrt(5)-1`
`rArr " " v lt (1)/(sqrt(5)-1)~~0.81m//s`
`rArr " " v lt 0.81 m//s`