Question

A particle moving along circular path of radius 1m completes 2(1/4)th revolution. Then displacement and distance of the particle are
A)
$\sqrt{3} and6 \frac{\pi2}{?} }$
B)
$\sqrt{2 \: and} 9 \frac{\pi}{2}$
​

Answer 1

Displacement at end of two revolutions is equal to zero. Now after next 1/4th revolution,

Using Pythagoras therorem

Displacement (see from fig) =

$\sqrt{ {1}^{2} + {1}^{2} } = \sqrt{1 + 1} = \sqrt{2}$

Distance = 2.25 × 2πr = 2.25 × 2π×1

= 4.5π = 9π/2

So, option b.

Answer 2

Answer:

The displacement of the given particle is √2 and distance covered by the given particle is 9π/2.

Explanation:

GIVEN : radius = 1meter

number of revolutions = 2 and 1/4th

We have to FIND : Displacement of the particle

Distance of the particle

First let us find the displacement

DISPLACEMENT: it is the shortest distance traveled by the object.

In the given question, the particle completes 2 and 1/4 th revolution.

so the displacement for the 2 revolutions is zero as the start and end point is same.

now let us calculate the displacement for the 1/4th revolution.

from the above picture if we consider the revolution is from A to C, by using the radius we can form a triangle.

The displacement will be the hypotenuse.

so by using Pythagorean law we can write that

${ab}^{2} + {bc}^{2} = {ac}^{2}$

we know that ab = bc = 1m

so,

${1}^{2} + {1}^{2} = {ac}^{2}$

$2 = {ac}^{2}$

√2 = ac

so, the displacement of the given particle is √2.

DISTANCE : it is the total area covered by the particle.

In the given question, the particle completes 2 and 1/4 th revolution.

distance covered in one revolution is equal to the circumference of the circular path way.

so distance of one revolution = 2πr

where r is the radius.

now it completed 2 and 1/4th revolution. so displacement will be

= 2πr + 2πr + (2πr/4 ) { as it completes only 1/4th revolution}

= 4π(1) + π(1) /2

= 9π/2

so, the distance covered by the particle is 9π/2.