Question

7.A player moves along the boundary of a square ground of side 50m in 200sec
The magnitude of displacement of the player at the end of 11 minutes 40seconds
from his initial position is
(a) 50m
(b) 150m
(c) 200m
(d) 50v2​

Answer 1

Explanation:

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Answer 2

$\bold \red{x = 50 \sqrt{2}}$

$\huge \star\huge \underline \bold \red{Explanation} \huge \star$

$\bold \red {11 \: min \: 40 \: sec=700 s}$

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 opposite vertex which we can find by using Pythagoras theorem

$x {}^{2} = 50 {}^{2} + 50 {}^{2}$

$x {}^{2} = 2500 + 2500 = 5000$

$x = \sqrt{5000}$

$x = 50 \sqrt{2}$