Question

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

Along with diagram ​

Answer 1

Solution :

First of all we will convert the total time of 2 minutes 20 seconds into seconds

Total time = 2 minutes 20 seconds

→ 2 × 60 seconds + 20 seconds

→ 120 seconds + 20 seconds

→ 140 seconds

Now,

Farmer made 1 round in 40 seconds

Rounds made by farmer in 140 seconds :

$\longrightarrow \sf \: \dfrac{1}{40} \times 140 \: rounds$

$\longrightarrow \sf \: \cancel\dfrac{140}{40} \: rounds$

$\longrightarrow \sf \: 3.5 \: rounds$

Thus,the farmer will made 3.5 rounds of the square field. If he started from the point A,then after the completion of 3 rounds he'll be at point A. In the next half round,He will be at B from A and B to C.So his final position will be at point C.

[ Refer to the attachment ]

Here,The Displacement of the farmer will be AC

Now,ABC is a right angled triangle with BC as base,AB as perpendicular and AC as hypotenuse

$\bf{By \: Pythagoras \: Theorem : }$

$\longrightarrow \sf \: ({AC})^{2} = ({AB})^{2} + ({BC})^{2}$

$\longrightarrow \sf \: ( {AC})^{2} = ({10})^{2} + ({10})^{2}$

$\longrightarrow \sf \: ( {AC})^{2} = 100 + 100$

$\longrightarrow \sf \: ( {AC})^{2} = 200$

$\longrightarrow \sf \: {AC} = \sqrt{200}$

$\longrightarrow \sf \: {AC} = 14.143 \: m$

Thus,The Magnitude of displacement of the farmer at the end of 2 minutes 20 seconds will be 14.143 metres.

Answer 2

Figure :

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ D }\put(5.5,5.8){ A }\put(9.8,5.8){ B }\put(9.8,9.1){ C }\put(4.5,7.5){ 10\:m }\put(7.2,5.3){ 10\:m }\put(10.1,7.5){ 10 \:m }\put(7.2,9.5){ 10\:m }\put(6,6){\line(1,0){3.7}}\put(6,9){\line(1,0){3.7}}\put(9.7,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(6,6){\line(5,4){3.75}}\end{picture}$

Given :

• Side of square = 10 m
• Time taken to complete 1 round = 40 s
• Time taken to cover the distance = 2 min 20 s

To find :

• Magnitude of displacement at the end of 2 minutes 20 s from initial position.

Solution :

Given that, side of square is 10 m.

Perimeter of square = 4 × side

: $\implies$ Perimeter of square = 4 × 10

: $\implies$ Perimeter of square = 40 m

Also, it is given that, he completes 1 round in 40 s.

$\therefore$ time taken = 40 sec and distance covered = 40 m

Now,

$\large{\boxed{\rm{Speed\:=\:\dfrac{Distance}{Time}}}}$

: $\implies$ Speed = $\dfrac{40}{40}$

: $\implies$ Speed = 1 m/s

$\rule{200}2$

It is alse given that it takes 2 min and 20 sec ( or 140 sec ) to cover the distance.

So, we can say that,

: $\implies$ Number of rounds = $\dfrac{140}{40}$

: $\implies$ Number of rounds = 3.5

Point to remember : Displacement is zero when object comes to it initial position.

$\therefore$ In three rounds, displacement is zero.

In the rest 0.5th part of round, farmer reaches the opposite end of diagonal of the square.

So, from Pythagoras theorem,

: $\implies$ Displacement ( AC ) = $\sf \sqrt{AB^2\:+\:BC^2}$

: $\implies$ Displacement ( AC ) = $\sf \sqrt{10^2\:+\:10^2}$

: $\implies$ Displacement ( AC ) = $\sf{\sqrt{200}}$

: $\implies$ Displacement ( AC ) = 14.14 m

$\therefore$ $\large{\boxed{\rm{Magnitude\:of\:displacement\:is\:14.14\:m}}}$