Question

Question 4. Pls answer

Answer 1

Answer:

$\bigstar$ Given:

→ Side of the square = 10 m

→ He completes 1 round in 40 seconds

$\bigstar$ To Find:

Displacement of the farmer at the end of $\sf 2 \;minutes\; 20 \;seconds$ from his initial position

$\bigstar$ Solution:

⇒ Side of the square = 10 m

The perimeter of the square,

$\implies \sf 10+10+10+10 = 40 m$

⇒ He completes 1 round in 40 s

So the the speed of the farmer can be calculated by using the formula,

$\implies \sf speed = \dfrac{distance}{time}$

$\implies \sf speed = \dfrac{40}{40}$

$\implies \sf speed = 1 \;m/s$

We need to calculate the displacement at 2 minutes 20 seconds

Converting time into seconds

$\implies \sf 2 \;minutes\; 20 \;seconds$

$\implies \sf 120\;s +20\;s = 140\;s$

Number of rounds completed in moving through 140 m

$\implies \sf rounds\; = \dfrac{140}{40}$

$\implies \sf rounds = 3.5$

Here, the farmer completes 3.5 rounds at 140 seconds

From this we can understand that,

So three times he returns back to the same place where he started.

Therefore, his displacement in three rounds in zero

Now the remaining distance he covered is $\sf 3.5\;m -3\;m = 0.5 \;m$

At 0.5 m from the starting point (after completion of three rounds)

The farmer reaches the diagonally opposite end of the square from his starting point.

Since,

Displacement is be shortest distance from the initial or starting point in a square

The diagonal will be the shortest distance

[As square all sides 10 cm]

Hence applying Pythagoras theorem,

(Hypotenuse)^2 = (base)^2 + (height)^2

Hypotenuse is the diagonal in square

$\implies \sf Displacement = \sqrt{10^2+10^2}\;\;m$

{Since it is a square, therefore the base and height are same}

$\implies \sf Displacement = \sqrt{200}\;\;m$

$\implies \sf Displacement = 10 \sqrt{2}\;\;m$

$\boxed{\boxed{\large{\sf Displacement = 10 \sqrt{2}\;\;m}}}$