Question

1 point
10.Aplayer moves along the boundary of a square ground of side 50 min
200 sec. The magnitude of displacement of the player at the end of 11
minutes 40 seconds from his initial position is
a.50m
O
b.150 m
O
C.200 m
0.50V2 m​

Answer 1

50 √2  m is the displacement of the player at the end of 11  minutes 40 seconds

Explanation:

Square ground of 50 m  sides

=> perimeter = 4 * 50 = 200 m

one round is completed in 200 Sec

=> Hence speed = 1 m/s

Distance Covered in 11 min 40 secs = 700 sec    = 700 m

700 = 200 + 200 + 200 + 100

Hence he has moved 100 m from initial position

50 m on one side  and 50 m at right angle

Hence Distance of final position from initial position =  √50² + 50²

= 50√2

50 √2  m is the displacement of the player at the end of 11  minutes 40 seconds

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