A farmer moves along the square field of side 10 m in 40 seconds what will be the magnitude of displacement the former at the end of 2 minutes 20 seconds
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The total distance covered in first round will be 40 m in time 40 seconds.
Also the final time is 2 min 20 seconds which is 140 seconds.
During each round( distance:40 m) the farmer comes back to its original position so the displacement is zero for every round or 40 metre.
So in 120 seconds even(3 rounds) its displacement will be zero.
But in remaining 20 seconds it will cover 2 side's distance. Suppose if the square field have corners named A,B,C and D respectively with A as the starting point;the farmer will be at corner C. As each side is 10 metres then using Pythagoras Theorem displacement AC will be (((10)^2)+((10)^2))^(1/2)
=(200)^(1/2)
=10((2)^(1/2))
So the displacement is 10× square root of 2
Also the final time is 2 min 20 seconds which is 140 seconds.
During each round( distance:40 m) the farmer comes back to its original position so the displacement is zero for every round or 40 metre.
So in 120 seconds even(3 rounds) its displacement will be zero.
But in remaining 20 seconds it will cover 2 side's distance. Suppose if the square field have corners named A,B,C and D respectively with A as the starting point;the farmer will be at corner C. As each side is 10 metres then using Pythagoras Theorem displacement AC will be (((10)^2)+((10)^2))^(1/2)
=(200)^(1/2)
=10((2)^(1/2))
So the displacement is 10× square root of 2
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