Question

A body moving along a circular path of radius 21 metres describe an angle of 120 degrees at the centre of the circle while moving from a to b find the distance and displacement of the body​

Answer 1

Answer:

Distance travelled = 44 m

displacement = 21√3 m

Explanation:

given that,

A body moving along a circular path of radius 21 metres

here,

radius of the circular track = 21 m

also given that,

angle describe at the centre of the circle while moving from a to b = 120°

now,

distance = length that the body covered

= length of the arc covered

length of the arc = 2πrθ/360°

here,

r = radius of the circular track

θ = angle described at the centre

putting the values,

distance = 2πrθ/360

2 × 22/7 × 21 × 120 /360

= 44 m

so,

length of arc = 44 m

so,

distance travelled = 44 m

now,

displacement = shortest distance between a and b

let the radius of the circular track be vector a and vector b

and hence by displacing the radius to get the resultant as the shortest distance between a and b

θ = 180 - 120

= 60°

ACCORDING TO THE FIGURE

r^-> - r^-> = displacement

and we know that,

when magnitude of two vectors rae equal then,

difference of the vectors = 2r sinθ/2

where,

r = magnitude of one vector

putting the values,

2 × 21 × sin120/2

42 × sin60

42 × √3/2

= 21√3 m

so,

displacement of the body

= 21√3 m

_________________

ANSWER:

distance travelled = 44 m displacement = 21√3 m

_________________

Answer 2

Answer:

The distance and displacement of the body are 44m and 36.37m

respectively.

Explanation:

Given that,

A body moving along a circular path of radius 21m describe an angle 120* at the center of the circle while moving from a to b.

If an arc of a circle subtends an angle $\theta$ at the center then the length of the arc is $l=2\pi r\times\frac{\theta}{360}$

Hence,the distance covered by body would be

$d=2\pi(21)\times \frac{120}{360}=44m$

The displacement is measured along the linear path(chord) between the point a and b and it is given by the relation,

$s=2rsin\frac{\theta}{2}$

Substituting the known values in above relation,we get

$s=2(21)sin{120}{2}=42sin60=21\sqrt{3}m=36.37m$

Therefore,

The distance and displacement of the body are 44m and 36.37m

respectively.

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