Question

a farmer moves along the boundary of a square field of side 10m in 40s. what be the magnitude of displacement of the farmer at the end of 2 minutes 20 second from his initial position?
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Answer 1

$\red\bigstar$ Explanation $\red\bigstar$

$\leadsto$ Given:-

Length the side of the square field = 10 m

Time taken to complete 1 round of the square field = 40s

$\leadsto$ Solution:-

Perimeter of the square field = 4 $\times$ 10m = 40 m

Time taken to complete 1 round of the square field = 40s

Time taken = 2 minutes 20 seconds = (2 $\times$ 60) seconds + 20 seconds = 140 seconds

Distance travelled in 140 seconds = Perimeter of the square $\times$ 140 seconds / 40 seconds [Using Unitary Method]

Distance travelled in 140 seconds = 40 $\times$ 140 seconds / 40 seconds = 140 m

We know that displacement is the shortest path between any two points

Number of rounds the farmer travelled in 140 seconds = $\dfrac {140}{40}$ [Using Unitary Method]

Number of rounds the farmer travelled in 140 seconds = 3.5 rounds

We also know that,

If we complete one round of a closed path then the displacement will be 0 m

Number of rounds the farmer travelled in 140 seconds = 3.5 rounds

Therefore,

For the three rounds the displacement will be 0.

Therefore,

The number of rounds for which the displacement of the farmer will be non zero in 140 seconds = 0.5 rounds

In 0.5 rounds the farmer will be at opposite point from the intial point

Therefore,

The displacement will be the diagonal of the square field.

Diagonal of the square field = √(10² + 10²) = 10√2

Therefore,

$\boxed{The\:displacement\:of\:the\:farmer\:=\:10√2m}$

Answer 2

Explanation:

distance covered by farmer in 40s = perimeter of square field

= 4 × side

= 4 × 10

= 40 m

speed of farmer = distance/time

= 40/40

= 1m/s

distance covered by farmer in 140sec ( 2min20sec)

= speed × time

= 1 × 140

= 140m

We know that displacement is the shortest distance between initial and final position

e know that displacement is the shortest distance between initial and final positionEverytime the farmer reaches the starting point displacement will become zero i.e. after every 40m

now for final position = 140/40

= 3.5

therefore the farmer will cover 3 complete round and 1/2 round (see figure in attachment)

therefore the farmer will cover 3 complete round and 1/2 round (see figure in attachment)displacement = shortest distance between A and C

using Pythagoras theorem

AC² = AB²+BC²

AC²= 10²+10²

AC²= 100+100

AC²= 200

AC = √200

AC = 10√2

MAGNITUDE OF DISPLACEMENT = 10√2m