Question

A particle is Moving in a circle of diametre 5m.Calculate the distance covered and the dusplacement when it completes 3 revolutions​

Answer 1

Step-by-step explanation:

Given,

Diameter of the particle = 5m

$\sf \: Radius \: = \frac{d}{2} = \frac{5}{2} = 2.5m$

To find

• Displacement when it completes 3 revolutions

Solution

$\sf \: Distance \: in \: 1 \: revaluation = 2\pi r\:$

$\sf \: Distance \: in \: 1 \: revaluation = 2 \times \frac{22}{7} \times 2.5$

$\sf \: Distance \: in \: 3\: revaluation = 3 \times \frac{22}{7} \times 2 \times 2.5$

( we could take the value of pi 3.14)

$\qquad \qquad \qquad \qquad \: \sf \twoheadrightarrow \: 3 \times 3.14 \times 5$

$\qquad \qquad \qquad \qquad \: \sf \twoheadrightarrow \:9.42 \times 5$

$\qquad \qquad \qquad \qquad \: \sf \twoheadrightarrow \: \red {\frak{47.10}}$

Hence, as the body come back to the same position the displacement is zero.

Answer 2

$\sf\red{Given :-}$

$\sf{Diametre = 5m}$

$\sf{radius = 2.5m}$

$\sf\red{Solution :-}$

$\sf{Distance \: in \: 1 \: revolution :-}$

$\sf{⟹ 2 × π × r}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf⟹ 2 × \frac{22}{7} \times 2.5$

$\sf{Distance \: in \: 3 \: revolution :-}$

$\sf{⟹ 3 × π × 2 × 2.5 }$

$\sf⟹ 3 × \frac{22}{7} \times 2 \times 2.5$

$\sf{ ⟹ 47.1}m$

★ Displacement is zero because it comes back to the position from where it was started.