Question

A man walks 40 meters towards north. Then turning to his right, he walks 50 meter. Then turning to his left, he walks 30 meters. Again he turns to his left and walks 40 meters.

How far is he from initial position?

Answer 1

Answer:

The required answer is 70.7 m

Step-by-step explanation:

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To find how far the man is from initial position, we have to find the displacement.

• The displacement is the shortest distance between the initial and final positions.

• It is a vector quantity as it has both magnitude and direction.

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$\rm \hat{i}$ is the unit vector along x - direction  

$\rm \hat{j}$ is the unit vector along y - direction

(1) The man initially walks 40 m towards North.

So, this displacement =  $\rm 40 \hat{j}$

(2) Then he walks 50 m towards his right. i.e., towards East

This displacement =  $\rm 50 \hat{i}$

(3) Turning to his left i.e., North he walks 30 m

Displacement = $\rm 30 \hat{i}$

(4) He walks 40 m turning to his left i.e., West

Displacement = $\rm 40(-\hat{i})$

The total displacement is the vector sum of all displacements.

 $\sf = 40\hat{j}+50\hat{i}+30\hat{j}+40(-\hat{i}) \\\\ \sf =40\hat{j}+50\hat{i}+30\hat{j}-40\hat{i} \\\\ \sf =10\hat{i}+70\hat{j}$

Magnitude of displacement :

$\tt = \sqrt{70^2+10^2} \\\\ \tt = \sqrt{4900+100} \\\\ \tt =\sqrt{5000} \\\\ \tt =10\sqrt{50} \\\\ \tt \approx 70.7 \ m$

The required answer is 70.7 m