Maganbhai moves along the boundary of a square field of side 20meter in 80seconds what will be the magnitude of displacement of maganbhai at end of 4min 40seconds from his initial position
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distance covered= 20m
total distance covered = 20 × 4 = 80m
time taken for 1 round = 80sec
total time taken = 4min 40 sec
= 4 × 60 + 40
= 240 + 40
= 280 sec
displacement = AB
(AB)² = (CD)² + (BD)²
(AB)²= 20² + 20²
= 400+400
= 800
AB = √(800) = 28.28
∴ Displacement = 28.28m
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1
Answer:
10 m because 20-4-6 =10 m
Explanation:
by velocity time relation v=u+at
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