Question

solve. A farmer moving along the boundary of a square field of side 12m complete its one round in 50 sec. find the magnitude of distance and displacement of the farmer at the end of 2min30sec from the initial position.

Answer 1

Hello Friend !

The answer to your query is given below :

Given, the side of the square = 12 m

Therefore, perimeter of the square = 12 × 4 = 48m

Farmer moves along the boundary in 50 sec.

Therefore, displacement after 2 min 30 sec = 2 × 60 sec + 30 sec = 150 sec = ?

Since in 50 sec farmer moves 48 m , therefore, in 1 sec distance cvered by farmer =  

⇒ $\frac{48}{50}$ = $\frac{24}{25} m$

Therefore, in 150 sec distance covered by the farmer = 

⇒ $\frac{24}{25} * 150$

⇒ 24 × 6 = 144 m .

This is the magnitude of distance covered by the farmer in 2 min 30 sec i.e., 150 sec

Now, number of rotations to cover 144 m along the boundary = $\frac{Total distance covered}{perimeter}$

⇒ no of rotations = $\frac{144}{48}$ = 3 rotations 

Since 3 complete rotations are made by the farmer in 2 min 30 sec , this means that the farmer returns to its initial position from where he started after 3 rotations.

Therefore the magnitude of displacement = 0 

This is your answer .

Hope this Helps !

Answer 2

Time taken for 1 round is 50 seconds
1 round = 50 seconds

Total rounds in 2 min 30 seconds (= 150 s) is
n = 150 seconds × (1 round / 50 seconds) = 3 rounds

Distance = 3 rounds × [(4 × 12 m) / round] = 144 m

Displacement = 0 [∵ After 3 rounds farmer is at its initial position]