Question

Mr. Hamza goes 30 meters North then turns right and walks 40 meters, then again turns right and walks 20 meters, then again turns right and walk 40 meters. How many meters is he from his original starting point?

Answer 1

Given :  Mr. Hamza goes 30 meters North then turns right and walks 40 meters, then again turns right and walks 20 meters, then again turns right and walk 40 meters.

To Find : How many meters is he from his original starting point

Solution:

                NORTH

WEST                           EAST

                 SOUTH

Mr. Hamza goes 30 meters North

North 30 m

Turns Right from North means  East    and walks 40 m

East  40 m

Turns Right from East means  South    and walks 20 m

South 20 m   =  North   -20 m

Turns Right from  South Means West and walks  40 m

West 40 m   =  East  = - 40 m

North = 30 - 20  = 10 m

East = 40 - 40 = 0 m

Hence Distance from starting point

is 10 M North  

Mr. Hamza  is  10 meters  from his original starting point

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Answer 2

Given:

Mr. Hamza goes 30 meters North then turns right and walks 40 meters, then again turns right and walks 20 meters, then again turns right and walk 40 meters.

To find:

Final position of the man w.r.t to initial position.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2,2){\vector(0,1){1.5}}\put(2,3.5){\vector(1, 0){2}}\put(4,3.5){\vector(0,-1){1}}\put(4, 2.5){\vector(-1,0){2}}\put(1,3){30m}\put(3,4){40m}\put(4.5,3){20m}\put(2.5,2){40m}\put(2,2){\circle*{0.2}}\end{picture}}$

[To see the diagram, please view this answer in website]

Calculation:

Vector Method:

The small black circle is the starting point.

Man first moves 30 m north ;

$\therefore \vec{d1} = 30 \: \hat{j}$

Then , man turns right and moves 40 m towards east ;

$\therefore\vec{d2} = 40 \: \hat{i}$

Then , man again turns right and moves 20m towards South;

$\therefore\vec{d3} = 20 \: ( - \hat{j})$

Finally, man turns right and moves 40 m towards west;

$\therefore \vec{d4} = 40 \: ( - \hat{i})$

So, net Displacement ;

$\vec{d}_{net} = \vec{d1} + \vec{d2} + \vec{d3} + \vec{d4}$

$= > \vec{d}_{net} = 30 \hat{j} + 40 \hat{i}+ 20( - \hat{j})+ 40 ( - \hat{i})$

$= > \vec{d}_{net} = 30 \hat{j} + 40 \hat{i} - 20 \hat{j} - 40 \hat{i}$

$= > \vec{d}_{net} = 30 \hat{j} - 20 \hat{j}$

$= > \vec{d}_{net} = 10 \hat{j}$

So, final position is 10 metres North from initial position.