Question

a player moves along the boundary of a square ground of side 50 metre in 200seconds.the magnitude of the displacement of the farmer and the end of 11minutes 40 seconds promise initial position is -
(A)- 50m
(B)- 150m
(C)- 200m
(D)- 50 route 2​

Answer 1

Answer:

11 min 40 sec=700 s

in 200 sec 4×50 m (boundary=perimeter)=200m

so in 700 s he move 700 m

so he make 3 and 1/2 rounds

the displacement will be the length of diagonal as he will be at opp vertex which will be by Pythagoras

${x}^{2} = {50}^{2} + {50}^{2} \\ {x}^{2} = 2500 + 2500 = 5000 \\ x = \sqrt{5000} = 50 \sqrt{2}$

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Answer 2

Answer:

11 min 40 sec=700 s

in 200 sec 4×50 m (boundary=perimeter)=200m

so in 700 s he move 700 m

so he make 3 and 1/2 rounds

the displacement will be the length of diagonal as he will be at opp vertex which will be by Pythagoras

\begin{gathered}{x}^{2} = {50}^{2} + {50}^{2} \\ {x}^{2} = 2500 + 2500 = 5000 \\ x = \sqrt{5000} = 50 \sqrt{2}\end{gathered}x2=502+502x2=2500+2500=5000x=5000=502

11 min 40 sec=700 s

in 200 sec 4×50 m (boundary=perimeter)=200m

so in 700 s he move 700 m

so he make 3 and 1/2 rounds

the displacement will be the length of diagonal as he will be at opp vertex which will be by Pythagoras

\begin{gathered}{x}^{2} = {50}^{2} + {50}^{2} \\ {x}^{2} = 2500 + 2500 = 5000 \\ x = \sqrt{5000} = 50 \sqrt{2}\end{gathered}x2=502+502x2=2500+2500=5000x=5000=502

*PLEASE MARK AS BRAINLIEST*