Question

2. A body covers the quadrant of a circle of radius 30m. Find the distance and
displacement.​

Answer 1

Answer:

Given:

A body covers the quadrant of a circle of radius 30 m.

To find:

Distance and displacement

Definitions:

• Distance is the total path length traversed by the object.
• Displacement is the shortest path between the starting and stopping point.

Diagram:

See the attached photo to understand better.

Calculation:

Distance :

Total circumference = 2πr

=> Total circumference = 2π (30)

=> Total circumference = 60π metres.

So distance will be ¼(total circumference) as the body travelled in a. quadrant.

$distance = \dfrac{1}{4} (60\pi) \\ = > distance = 15\pi \: metres$

Displacement :

Shortest path can be calculated based on Pythagoras Theorem .

$displacement = \sqrt{ {30}^{2} + {30}^{2} }$

$= > displacement = 30 \sqrt{2} m$

Answer 2

Given,

A body covers the quadrant of a circle of radius 30m.

To find: the distance and displacement.

we know that,

Distance =Actual path covered by a body irrespective of direction.

Displacement =The short path between initial position and final position is called displacement.

$\mathbf \red{To \: Calculated:-}$

Distance of circle = Total circumference = 2πr(radius is given)

=> Total circumference = 2π (30)

=> Total circumference = 60π metres.

So ,distance will be ¼(total circumference) as the body travelled.

distance = 1/4 (60π)

=15πm

Now,

$\mathbf \red{Using \: pythagoras \: therome }$

displacement =

$\sqrt{r {}^{2} } + \sqrt{r {}^{2} }$

$\mathbf \red{putting \: the \: value \: of \: radius}$

$\sqrt{30} {}^{2} + \sqrt{30 {}^{2} }$

$displacement = 30 \sqrt{2}$