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?
Answers
the perimeter is 40m
so his speed= 1m/s
time=2m20s=140s
distance covered= 140m
no. of rounds= 140/40=2.5
so he completes 2 rounds so displacement is 0
but .5 round means 2 side so net displacement is 10√2m (see attachment for clear understanding)
First of all we will convert the total time of 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 s, number of rounds made = 1
So , In 140 s, number of rounds made = 140 ÷ 40 = 3.5
Thus , the farmer will make three and half rounds , i.e. , 3.5 rounds of the square field. If the farmer starts from position A, then 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 angle 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
AC = √200
AC = 14.143 m
Thus , the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds will be 14.143 metres.
