Question

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

Answer 1

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$given \\ \\ side \: of \: \: square \: field = 10m \\ perimeter = 10 \times 4 = 40m \\ \\ farmer \: moves \: along \: the \: field \: in \: \: 40sec \\ displacement \: after \: 2m \: 20s = 2 \times 20 + 20 = 140s \\ since \: in \: 40sfarmer \: moves \: 40m \\ therefore \: in \: 1s \: distance \: covered \: by \: farmer \: = \frac{40}{40} = 1m \\ therefore \: in \: 140s \: the \: farmer \: covers \: 140m \\ \\ now \: the \:number \: of \: rotation \: to \: cover \: 140 \: along \: the \: boundary \: = \frac{total \: distnce }{total \: time}$