Question

A farmer moves along the boundary of a field of side 10 meters in 40 seconds . What will be the magnitude of displacement of the farmer at the end of 20 minutes 20 seconds from his initial position

Answer 1

$\huge\underline\boxed\bold\red{Answer}}$

Total distance 40m in 40 sec.

Total time taken by the farmer = 2 min 20 sec. =140 seconds.

Total rounds completed= 140/40=3.5 rounds

That means if the farmer starts from point A of the square field, he reaches point C.

Therefore, displacement is AC.

Assume ABC is a right angled triangle.

Therefore,

$AC^{2}=AB^{2} +BC^{2}$

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

$AC^{2}=100+100$

$AC^{2}=200$

$AC=(200)^{\frac{1}{2} }$

$AC^{2} =10\times (2)^{\frac{1}{2} }$

$\underline\boxed\bold\blue{Hope it helps :)}}$