Question

A boy moves 4 metres towards east then turns left by 135 degrees and travels for 2 root 2 metres. From there he takes 90° degrees and moves for about 2 root 2 metres. After that he travels 8 metres towards east and then 1 metre towards north. Find the magnitude of displacement of boy.

Answer 1

Answer:

13m

Step-by-step explanation:

Answer 2

Answer:

202.88m

Step-by-step explanation:

Given:

The boy walks 4 metres eastward. then makes a 135° left turn and moves 2 metres.

To find:

Magnitude

Soluiton:

Any change in a point or object's position is referred to as displacement. It is a vector quantity since it has a direction and a magnitude. An arrow pointing from the initial location to the ending position is used to represent it. As an illustration, suppose that an object is transported from location A to position B, then its position is altered. Displacement is the shortest distance between positions A and B. The displacement formula will be covered in this article.

In order to answer this question, we must first understand what displacement is and how it is expressed mathematically. To further our understanding, we will use a problem to determine the magnitude of displacement. We shall also see the fundamentals of motion equations.

AB=

$=2(cos135^{o}i+sin135^{o}j )\\=2(\frac{-1}{\sqrt{2} }i+\frac{1}{\sqrt{2} }j)\\ =-\sqrt{2}i +\sqrt{2} j$

He moves for roughly 212 m after turning 90 ° right from point B.

$BC=212(-cos45^{o}i-sin45^{o}j)\\ =-106\sqrt{2}i-106\sqrt{2}j$

He then walks 8 metres eastward after that

$CD=8i$

and finally goes 1 m in the direction of north,

$DE=1j$

Displacement = $OE= OA + AB + BC + CD + DE$

Substituting the values,

$= 4i - \sqrt{2}i + \sqrt{2}j - 106\sqrt{2}i - 106\sqrt{2} j + 8i + 1j\\=(12-107\sqrt{2})i+(1-105\sqrt{2})j\\=\sqrt{((12-107\sqrt{2})i)^{2}+((1-105\sqrt{2})j)^{2}} \\=202.88m$

202.88m

Step-by-step explanation:

Given:

The boy walks 4 metres eastward. then makes a 135° left turn and moves 2 metres.

To find:

Magnitude

Soluiton:

Any change in a point or object's position is referred to as displacement. It is a vector quantity since it has a direction and a magnitude. An arrow pointing from the initial location to the ending position is used to represent it. As an illustration, suppose that an object is transported from location A to position B, then its position is altered. Displacement is the shortest distance between positions A and B. The displacement formula will be covered in this article.

In order to answer this question, we must first understand what displacement is and how it is expressed mathematically. To further our understanding, we will use a problem to determine the magnitude of displacement. We shall also see the fundamentals of motion equations.

AB=

$=2(cos135^{o}i+sin135^{o}j )\\=2(\frac{-1}{\sqrt{2} }i+\frac{1}{\sqrt{2} }j)\\ =-\sqrt{2}i +\sqrt{2} j$

He moves for roughly 212 m after turning 90 ° right from point B.

$BC=212(-cos45^{o}i-sin45^{o}j)\\ =-106\sqrt{2}i-106\sqrt{2}j$

He then walks 8 metres eastward after that

$CD=8i$

and finally goes 1 m in the direction of north,

$DE=1j$

Displacement = $OE= OA + AB + BC + CD + DE$

Substituting the values,

$= 4i - \sqrt{2}i + \sqrt{2}j - 106\sqrt{2}i - 106\sqrt{2} j + 8i + 1j\\=(12-107\sqrt{2})i+(1-105\sqrt{2})j\\=\sqrt{((12-107\sqrt{2})i)^{2}+((1-105\sqrt{2})j)^{2}} \\=202.88m$

202.88m

Step-by-step explanation:

Given:

The boy walks 4 metres eastward. then makes a 135° left turn and moves 2 metres.

To find:

Magnitude

Soluiton:

Any change in a point or object's position is referred to as displacement. It is a vector quantity since it has a direction and a magnitude. An arrow pointing from the initial location to the ending position is used to represent it. As an illustration, suppose that an object is transported from location A to position B, then its position is altered. Displacement is the shortest distance between positions A and B. The displacement formula will be covered in this article.

In order to answer this question, we must first understand what displacement is and how it is expressed mathematically. To further our understanding, we will use a problem to determine the magnitude of displacement. We shall also see the fundamentals of motion equations.

AB=

$=2(cos135^{o}i+sin135^{o}j )\\=2(\frac{-1}{\sqrt{2} }i+\frac{1}{\sqrt{2} }j)\\ =-\sqrt{2}i +\sqrt{2} j$

He moves for roughly 212 m after turning 90 ° right from point B.

$BC=212(-cos45^{o}i-sin45^{o}j)\\ =-106\sqrt{2}i-106\sqrt{2}j$

He then walks 8 metres eastward after that

$CD=8i$

and finally goes 1 m in the direction of north,

$DE=1j$

Displacement = $OE= OA + AB + BC + CD + DE$

Substituting the values,

$= 4i - \sqrt{2}i + \sqrt{2}j - 106\sqrt{2}i - 106\sqrt{2} j + 8i + 1j\\=(12-107\sqrt{2})i+(1-105\sqrt{2})j\\=\sqrt{((12-107\sqrt{2})i)^{2}+((1-105\sqrt{2})j)^{2}} \\=202.88m$

Magnitude of the displacement is equal to the distance only when object is in

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