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 20 seconds from his initial position?

Answer 1

Total time t = 2 min 20 sec = 140 sec

Now, perimeter of field is
d = 4×l = 4×10 = 40 m

Given that farmer moves along the field once every 40 sec
Speed = Distance/Time
So, v = 40/40 = 1 m/s

That means farmer covers 1 metre in every 1 second

In 2 min, there are 120 sec. In this time, farmer will have completed 3 complete rounds (which is 120 m distance), and is back to initial position.

In the next 20 s, the farmer will move
10 m along one side, and then 10 m along other side.

So, finally farmer is diagonally opposite to where he started.

Apply Pythagoras Theorem to the square, and you get displacement as
10√2 metres

Answer 2

Answer:

Total time t = 2 min 20 sec = 140 sec

Now, perimeter of field is

d = 4×l = 4×10 = 40 m

Given that farmer moves along the field once every 40 sec

Speed = Distance/Time

So, v = 40/40 = 1 m/s

That means farmer covers 1 metre in every 1 second

In 2 min, there are 120 sec. In this time, farmer will have completed 3 complete rounds (which is 120 m distance), and is back to initial position.

In the next 20 s, the farmer will move

10 m along one side, and then 10 m along other side.

So, finally farmer is diagonally opposite to where he started.

Apply Pythagoras Theorem to the square, and you get displacement as

10√2 metres

Explanation: