Question

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

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Displacement=10\sqrt{2}\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Side \: of \: square = 10 \: m \\ \\ \tt: \implies Time \: taken \: to \: move \: along \: boudary = 40 \: sec \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Magnitude \: of \: Displacement \: after \: 2 \:min \: 20 \: sec = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies Perimeter \: of \: square = 4 \times side \\ \\ \tt: \implies Perimeter \: of \: square = 4 \times 10 \\ \\ \tt: \implies Perimeter \: of \: square =40 \: m \\ \\ \bold{For \: Speed : } \\ \tt: \implies Speed = \frac{Distance}{Time} \\ \\ \tt: \implies Speed = \frac{40}{40} \\ \\ \tt: \implies Apeed = 1 \: m/s \\ \\ \bold{For \: Distance \: travelled \: in \: 2 \: min \: 20 \: sec} \\ \tt: \implies Distance =Speed \times Time \\ \\ \tt: \implies Distance =1 \times 140 \\ \\ \tt: \implies Distance = 140 \: m \\ \\ \bold{For \: Displacement} \\ \tt: \implies One \: rotation = 40 \: m \\ \\ \text{So,\: after \: 3 \: rotation = 120 \: m } \\ \text{Displacement \: will \: be \: zero} \\\\ \tt{Ground\:is\:square\:so,}\\ \\ \tt: \implies Displacement =\sqrt{10^{2}+10^{2}}\\ \\ \green{\tt: \implies Displacement = 10\sqrt{2} \: m}$

Answer 2

$\tt{\huge{\red{Hello!!! }}}$

Question :

A farmer moves along the 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 and 20 seconds from his initial position.

Solution :

$\tt{\orange {Step-By-Step-Explaination \::- }}$

The side of the square field is 10 m.

Hence the perimeter of the field is 40 m.

The farmer can complete one round in 40 s.

Hence speed of the farmer = Distance / Time = 40 m / 40 s = 1 m/s

Competing 140 rounds, the farmer covers 140 / 40 rounds i.e, 3.5 rounds.

After completing three rounds the farmer reaches his initial position.

Hence Displacement = 0.5 round =>

$\tt{\purple{\leadsto{Displacement = \sqrt { {10}^2 + {10}^2 } = 10 \sqrt {2} \: m. }}}$