Question

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

Plz answer fast with full explanation and steps.

Answer 1

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

$\bold\green{\underline{\underline{Solution}}}$

$\bold{2min \ and \ 20sec=140sec}$...Given

$\bold{Side \ of \ square \ field=10m}$...Given

$\bold{Farmer \ take \ 40s \ to \ move \ along \ one \ side}$...Given

Lets find how many such sides can be cover

$\bold{\frac{140}{40}}$

Farmer can cover three such sides of square with 20 seconds left.

$\bold{Distance \ cover \ in \ 1sec= \frac{10}{40}}$

$\bold{Distance \ covere \ in \ 1sec=0.25m}$

$\bold{Remaining \ distance \ to \ cover \ on \ fourth \ side}$

$\bold{5m}$

In a right angled triangle form,

By Pythagorus theorem

(Refer the attachment)

$\bold{Displacement^{2}=10^{2}+5^{2}}$

$\bold{Displacement^{2}=100+25}$

$\bold{Displacement^{2}=125}$

On taking square root of both sides

$\bold{Displacement=11.18m}$

$\bold\red{Therefore \ displacement \ of \ farmer \ from}$

$\bold\red{the \ initial \ position \ is \ 11.18 \ metre.}$