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. A farmer moves along the boundary of a square field of side10 m in 40 s. What will be themagnitude of displacement of thefarmer at the end of 2 minutes 20 |
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Answer» Solution : FIRST of all convert the given time 2 minutes 20 seconds into seconds. Total Time = 2 minutes 20 seconds → 2 × 60 seconds + 20 seconds → 120 seconds + 20 seconds → 140 seconds Now, In 40 seconds,Round made = 1 So, In 140 seconds,Round made = (1/40)*140In 140 seconds = 3.5 rounds Thus,the farmer will make three and half rounds of the square field. If the farmer starts from position A,then after completion of 3 rounds, he'll be at STARTING position A. But in the next half round,Farmer will MOVE from A to B,and B to C,so that his final position will be at C. Thus,the net DISPLACEMENT of farmer will be AC. Now ABC is a right ANGLED triangle in which AC is the hypotenuse. [Refer to the attachment] So,∴ The magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds will be 14.143 metres. |
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