Question

Plz answer quickly, its urgent.
A particle moves over three quarters of a circle of radius 'r'. What is the magnitude of its displacement?

Answer 1

Answer:

This is my answer

Explanation:

Answer 2

Explanation:

Let a be the initial and b be the final position (see in attachment)

Then the displacement covered by body is equal to the shortest distance between a and b.

In right ∆ AOB,

OA = OB = r

using pythagoreas theorm,

⭐ $\boxed{\sf</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{ </strong><strong>AB</strong><strong>= \sqrt{ {</strong><strong>OA</strong><strong>}^{2} + {</strong><strong>OB</strong><strong>}^{2} } </strong><strong>}}$

$\implies$ $\bold{</strong><strong>A</strong><strong>B</strong><strong> </strong><strong>=</strong><strong> </strong><strong> </strong><strong>\sqrt{ {r}^{2} + {r}^{2} } </strong><strong>}$

$\implies$ $\bold{</strong><strong>A</strong><strong>B</strong><strong> </strong><strong>=</strong><strong> </strong><strong> \sqrt{2 {r}^{2} } </strong><strong>}$

$\implies$ $\bold{</strong><strong>A</strong><strong>B</strong><strong> </strong><strong>=</strong><strong> r \sqrt{2} </strong><strong> </strong><strong>units</strong><strong> </strong><strong>}$

hence, the magnitude of displacement of three quarter of a circle of radius r is $\sf</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{\underline{</strong><strong> r \sqrt{2} </strong><strong>}}$