Solveeit Logo
Login

Question

Question: A man wants to reach from A to the opposite corner of the square C (Figure). The sides of the square...

A man wants to reach from A to the opposite corner of the square C (Figure). The sides of the square are 100 m each. A central square of 50 m × 50 m is filed with sand. Outside this square, he can walk at a speed 1 m s11 \mathrm {~m} \mathrm {~s} ^ { 1 } . In the central square, he can walk only at a speed of vms−1\mathrm { v } \mathrm { m } \mathrm { s } ^ { - 1 } (v < A). 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?

A

0.18 m s−10.18 \mathrm {~m} \mathrm {~s} ^ { - 1 } (0.18 m/s0.18 \mathrm {~m} / \mathrm { s })

B

0.81 m s−10.81 \mathrm {~m} \mathrm {~s} ^ { - 1 } ()

C

0.5 m s−10.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }( 0.5 m/s0.5 \mathrm {~m} / \mathrm { s } )

D

0.95 m/s0.95 \mathrm {~m} / \mathrm { s }s)

Answer

0.81 m s−10.81 \mathrm {~m} \mathrm {~s} ^ { - 1 } ()

Explanation

Solution

AC=(100)2+(100)2=1002 m\mathrm { AC } = \sqrt { ( 100 ) ^ { 2 } + ( 100 ) ^ { 2 } } = 100 \sqrt { 2 } \mathrm {~m}

AP=AC−PQ2=1002−5022\mathrm { AP } = \frac { \mathrm { AC } - \mathrm { PQ } } { 2 } = \frac { 100 \sqrt { 2 } - 50 \sqrt { 2 } } { 2 }

=252= 25 \sqrt { 2 } m

QC=AP=252 m\mathrm { QC } = \mathrm { AP } = 25 \sqrt { 2 } \mathrm {~m}

RC=AR=2510 m\mathrm { RC } = \mathrm { AR } = 25 \sqrt { 10 } \mathrm {~m}

Consider the straight line path APQC through the sand. Time taken to go from A to C via this path

=252+2521+502v=502[1v+1]= \frac { 25 \sqrt { 2 } + 25 \sqrt { 2 } } { 1 } + \frac { 50 \sqrt { 2 } } { \mathrm { v } } = 50 \sqrt { 2 } \left[ \frac { 1 } { \mathrm { v } } + 1 \right]

The shortest path outside the sand will be ARC. Time taken to go from A to C via this path

=2510+25101=5010= \frac { 25 \sqrt { 10 } + 25 \sqrt { 10 } } { 1 } = 50 \sqrt { 10 }