Question

A person moves along the boundary of a square park of side 10 m in 40 s. What will be the magnitude of displacement and distance of the person at the end of 100 seconds?

Answer 1

Answer:

The magnitude of displacement of the person is 14.14m

The distance of the person at the end of 100 seconds is 100m

Explanation:

Given that, A person moves along the boundary of a square park of side 10 m in 40 seconds;

A person an complete one round in 40 seconds

Now, the speed of a person is

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

v=40/40

v=1m/s

So, the distance covered in 100 seconds is

d=v x t

d=1 x 100

d=100m

Hence, the distance covered in 100 seconds is 100m

In a 4 rounds, the displacement is zero.

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

The displacement is diagonal of the park (square)

Using the Pythagorean theorem, we get

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

D=$\sqrt{100+100}$

D=10$\sqrt{2}$

D=14.14m

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

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