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
Answers
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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