Question

1.
A particle is moving along a circle such that it
completes one revolution in 40 seconds. In 2
minutes 20 seconds, the ratio us!
stip displacement is
distance
E
(2)​

Answer 1

Answer:

Given:

Particle completes pne revolution in 40 seconds.

To find:

Ratio of displacement and distance at a time of 2minutes 20 seconds.

Concept:

Distance is the total path length travelled by a body. It is always positive and a scalar quantity.

Displacement is a vector Quantity representing the shortest length between starting and stopping point.

Calculation:

For each revolution, time taken = 40 sec

In 2 minutes 20 secs (i.e. 140 secs) , the number of revolutions that will be made by that particle will be :

3 Full Revolutions and 1 half Revolution.

So displacement be "d"

$\boxed{ \bold{d = 2r}}$

Let distance be "s"

$s = 3 \times (2\pi r) + 1(\pi r)$

$\boxed{ \bold{ s = 7\pi r}}$

So ratio will be :

$\boxed{d \: : \: s = 2r : 7 \pi r = 2 : 7\pi}$

So final answer :

$\boxed{ \huge{ \red{ \bold{d : s = 2 : 7\pi}}}}$

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Question-}}}}}}$

A particle is moving along a circle such that it

completes one revolution in 40 seconds. In 2

minutes 20 seconds, the ratio is?

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Solution:}}}}}}$

Given:

Particle completes pne revolution in 40 seconds.

To find:

Ratio of displacement and distance at a time of 2minutes 20 seconds.

For every revolution,

Time taken = 40 Seconds

In 2 minutes 20 seconds or 140 seconds,the number of revolutions would be 3 Full revolutions and a half revolution

Let the displacement be D

D = 2r

Let distance be P

P = 3 × (2πr) + 1 (πr)

P= 7πr

So, the ratio obtained would be :-

D : P = 2r : 7πr = 2 : 7π

$\huge{\\ 2 \: : \: 7\pi \: }$