A farmer moves along the boundary of a square field of side 10 m in 40 s. what will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

hello this question we have given that a farmer moves along the boundary of a square field of side 10 m in 40 seconds so what will the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds in this question we have given a square field this is a square field and it is given in the question that a farmer moves along the boundary of a square field of 10 m to the side is given to us that is 10 M 10 metre and in 40 seconds recover the one round in 40 second we have to find the magnitude of the Palmer at the end of 2 minutes 20 seconds so now in this question we have you the time for one round time for background which is equal to 40 second in the total time is 2 minut 20 second and which will be equal to 2 into 6027 time here is equal to 140 2nd and come here we can calculate the number of town the number of ground that is an equal to the total time over time for one round equal to 3.5 to the total amount is 3.5 if a farmer start from this point let this point is a b then point start from this point from a then after he complete third round then in the last point Hyderabad the star starting from A so it's half his 3.5 down will finish in C point so have a displacement will be equal to displacement is equal to so we have to find the magnitude that what will be the magnitude of the farmer this will be the magnitude of the farmer at the end of 2 minutes 20 seconds because he has completed 3.3 down he completed here and point 5 will be here at sea for the displ...

The correct option is B 10 √ 2 m In the given figure A B C D is a square field of side 10 m Time for one round is 30 s and total time of motion is 1 min 45 sec = 105 s So, number of round completed is 105 30 = 3.5 Thus, if farmer starts from A, he will complete 3 rounds at A. In the last 0.5 round starting from A, he will finish at C. ∴ Displacement A C = √ ( A B ) 2 + ( B C ) 2 = √ 10 2 + 10 2 = 10 √ 2 m

Side of the square field = Perimeter of the square = According to question, He completes 1 round in . Speed of the farmer = Distance covered in = Now, Number of round tripscompleted travelling = We know, in 3 round tripsthe displacement will bezero. In round, the farmer will reach diametrically opposite to his initial position. Displacement =

Total distance covered by the farmer 200 m in 200 sec. Speed of farmer = 200 m 200 s = 1 m s − 1 Total time taken by the farmer = 11 min 40 sec. = 700 seconds Total rounds completed= 700/200=3.5 rounds That means if the farmer starts from a point on the square field, he will reach a point diagonally opposite to that since after 3 rounds where he comes back to the same point he goes half a round which is a point diagonally opposite to the starting point. Therefore, displacement is length of the diagonal of the square. Let the length of diagonal be l Therefore, AC2= AB2 +BC2 l 2 = 10 2 + 10 2 ∴ l = √ 100 + 100 ∴ l = 10 √ 2 m