Question

A Girl moves along the boundary of a square field of side 20metres in 80seconds. What will be the magnitude of displacement at the end of 200 seconds. Also Calculate the velocity.

Answer 1

Answer:

$20\sqrt{2}$

Explanation:

The girl completes 1 round in 80 seconds.

∴ She will complete $\frac{200}{80}$ rounds in 200 seconds.

i.e. 2.5 rounds.

So, after 2 complete rounds the girl's displacement will become zero.

And for the next half round, she will be on the opposite corner of the field.

So by applying the pythagoras theorem,

$Displacement = \sqrt{a^{2} +b^{2}  }$

where a and b are the sides of the field travelled by her.

∴ $Displacement = \sqrt{20^{2}+20^{2} }\\\\= \sqrt{2*20^{2}}\\ = 20\sqrt{2}$ metres.