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 ?

Explanation

First of all we will convert the total time of 2 minutes 20 seconds into seconds.
Total time = 2 minutes 20 seconds
`" "= 20 xx 60` seconds + 20 seconds
`" "` = 120 seconds + 20 seconds
`" "` = 140 seconds
Now, In 40 s, number or rounds made = 1
So, In 140 s, number of rounds made = `(1)/(40)xx140`
`" "` = 3.5

Thus, the farmer will make three and a half rounds (3.5 rounds) of the square field. If the farmer starts from position A (see Figure), then after three complete rounds, he will reach at starting position A. But in the next half round, the 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 the farmer will be AC. Now, ABC is a right angled triangle in which AC is the hypotenuse.
So,
`" "(AC)^(2)=(AB)^(2)+(BC)^(2)`
`" "(AC)^(2)=(10)^(2)+(10)^(2)`
`" "(AC)^(2)= 100+100`
`" "(AC)^(2)= 200` ltBrgt `" "AC=sqrt(200)`
`" "` AC = 14.143 m
Thus, the magnitude of displacement of the farmer at the end of 2 minutes and 20 seconds will be 14.143 metres.