Question

a person moves along the boundary of a square of side 10m in 40s. What will be the magnitude of displacement of the person at the end of 3 minutes?
pls tell the answer with steps...fast

Answer 1

Answer:

your answer is 14.14meters displace of a person.

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=td

v =\dfrac{40}{40}v=4040

v = 1 m/sv=1m/s

So, The distance covered in 180 sec is

d = v\times td=v×t

d = 1\times180d=1×180

d = 180\ md=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=(10)2+(10)2

D=\sqrt{100+100}D=100+100

D= 10\sqrt{2}D=102

D=14.14\ mD=14.14 m

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