Question

. 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​

Answer 1

Answer:

Solution :

First of all convert the given time 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 seconds,Round made = 1

So,

In 140 seconds,Round made = (1/40)*140

In 140 seconds = 3.5 rounds

Thus,the farmer will make three and half rounds of the square field. If the farmer starts from position A,then after completion of 3 rounds, he'll be at starting position A. But in the next half round,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 farmer will be AC. Now ABC is a right angled triangle in which AC is the hypotenuse.

[Refer to the attachment]

So,

$=> (AC^{2} )=(AB^{2})+(BC^{2})$

$=> (AC^{2} )=(10^{2} )+(10^{2} )$

$=> (AC^{2} )=100+100$

$=> (AC^{2} )=200$

$=> AC=\sqrt{200}$

$=> AC =14.43 m$

∴ The magnitude of the displacement of the farmer at the end of 2 minutes 20 seconds will be 14.143 metres.