Question

A farmer moves along the boundary of a square field of side 10m in 40seconds.What will be the magnitude of displacement of the farmer at the end of 2minutes and 20seconds.​

Answer 1

Total distance covered 10m in 40sec

Total time taken= 2min 20sec= 140sec

Total round completed = 140/40=3.5 rounds

If an object moves from A to D, the shortest path will be from A to C. So displacement is AC, which we can find by phythogoras theorem.

AC^2= AB2 +BC^2 

AC^2= (10)^2 +(10)^2

AC^2= 100+100

AC2=200

AC = √ 200

AC= 10√ 2

Hence, displacement will be 10√2

Hope it helps you!

Answer 2

Answer:

Displacement = 14.142 m/s

Explanation:

Given-

• Side of square = 10 m
• Time taken to take one round around the field= 40 secs
• Total time taken = 2mins 20s

ㅤ

To Find-

• Second Initial position.

ㅤ

Solution -

Converting mins to s =

2 mins 20s = 2(60)+20= 120+20 = 140 s

ㅤ

Rounds taken in 140 s = 140÷40 = 3.5 rounds

ㅤ

[make a square ABCD and mark every side as 10m for better understanding]

ㅤ

By observing the question, we get to know that the intial point(displacement) is the diagonal of the field.

ㅤ

So now finding the diagonal through Pythagoras theorem -

Hypotenuse² = height² +base²

${x}^{2} = {10}^{2} + {10}^{2} \\ \\ {x}^{2} = 100 + 100 \\ \\ {x}^{2} = 200 \\ \\ {x}^{2} = 10 \sqrt{2} \\ \\ x = 14.142 \: m {s}^{ - 1}$

ㅤ

Displacement = 14.142 m/s