Question

a person moves along the boundary of a square Park of side 10 metre in 40 seconds. what will be the magnitude of displacement of the person at the end of 3 minutes ?

Answer 1

a person completed 4.5 rounds in 3 minutes.
Magnitude of displacement=
$= {10}^{2} + {10}^{2}$
$= 100 + 100$
$= 200$
$= \sqrt{200}$
$= 10 \sqrt{2} m$

Answer 2

Answer:

The magnitude of displacement of the person is 14.14 m.

Explanation:

Given that,

A person moves along the boundary of a square Park of side 10 meter in 40 seconds.

A person complete one round in 40 second.

Now, the speed of a person is

$v = \dfrac{d}{t}$

$v =\dfrac{40}{40}$

$v = 1 m/s$

So, The distance covered in 180 sec is

$d = v\times t$

$d = 1\times180$

$d = 180\ m$

In a 4 rounds, the displacement is zero.

In 0.5 round, a person reaches the diagonally opposite end of the square from his initial point.

Now, The displacement is diagonal of the park

Using Pythagorean theorem, we get

$D= \sqrt{(10)^2+(10)^2}$

$D=\sqrt{100+100}$

$D= 10\sqrt{2}$

$D=14.14\ m$

Hence, The magnitude of displacement of the person is 14.14 m.