Question

Calculate the distance and displacement of a body after completing ¾ part of the circular path.

Answer 1

Answer:

$Distance=\frac{3\pi r}{2}, \\ Displacement=r\sqrt{2}$

Explanation:

Let the radius of the circular path be r

=>|A|=|B|=r

Where A and B are the position vectors of the body at time t=0 s and T s respectively(refer to attachment)

Distance covered =

Length of major arc PP'

= $\frac{3}{4}×2\pi r\\ =\frac{3\pi r}{2}$

Angle between A and B

$=\theta \\ =[(1-\frac{3}{4})×360]°\\ =[\frac{1}{4}×360]°\\ =90°$

Displacement between A and B

=$\bold{| \triangle P| }$

$=\bold{ |B|-|A| }\\ =\bold{ |B|+(-|A|) } \\ =\bold{\sqrt{(-|A|)^{2}+(|B|)^{2}}}\\ =\sqrt{r^{2}+r^{2}}\\ =\sqrt{2r^{2}}\\ =r\sqrt{2}$

Direction of displacement

$=\alpha\\ =tan^{-1}\bold{(\frac{|B|}{|A|})}\\ =tan^{-1}(\frac{r}{r})\\ =tan^{-1}1\\ =45°\: above \: the\: position\: vector\: \bold{A}$

Answer 2

Answer:

Let, radius of the circular path = r units.

(3 / 4)th of the circular path = traversing through an angle 270°.

Distance travelled = (3 / 4) * (2 * π * r) units = (1.5 * π * r) units.

Remaining angle to cover one complete round = (360° - 270°) = 90°.

Displacement = {2 * r * sin (90° / 2)} units = {(√2) * r} units at an angle 45° backward.