Question

Easy Physics Question!!!

A person facing North walks 3 km, turn left and walks 4 km. Again he walks 3 km after turning right, then he turn to right and walks 12 km. What is his displacement from the starting point and which direction is he facing!?

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Answer 1

DIAGRAM:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(0,3)\qbezier(0,3)(0,3)(-3,3)\qbezier(-3,3)(-3,3)(-3,6)\qbezier(-3,6)(-3,6)(9,6)\qbezier(0,6)(0,6)(0,0)\qbezier(9,6)(9,6)(0,0)\put(0,-0.2){\sf A}\put(-3.2,3.2){\sf D}\put(0.4,2.8){\sf E}\put(0,6.2){\sf B}\put(-3.2,6){\sf F}\put(8.9,5.8){\sf C}\end{picture}$

Solution:-

Here,

• AE=3km
• BE=3km
• DE=4km
• DF=3km
• CF=12km

We can see

• DE||BF

Its a rectangle so

• BF=4km

Now

$\\ \sf\longmapsto BC=CF-BF$

$\\ \sf\longmapsto BC=12-4$

$\\ \sf\longmapsto BC=8km$

And

$\\ \sf\longmapsto AB=AE+BE$

$\\ \sf\longmapsto AB=3+3=6km$

Here

• $\sf \triangle ABC$is Right angled.

Where

• Perpendicular=P=BC=8km
• Base=B=AB=6km
• Hypontenuse=H=AC=?

According to Pythagorean theorem

$\\ \sf\longmapsto H^2=P^2+B^2$

$\\ \sf\longmapsto H=\sqrt{P^2+B^2}$

$\\ \sf\longmapsto AC=\sqrt{8^2+6^2}$

$\\ \sf\longmapsto AC=\sqrt{64+36}$

$\\ \sf\longmapsto AC=\sqrt{100}$

$\\ \sf\longmapsto AC=10km$

Hence Displacement=10km.

Answer 2

EXPLANATION:-

Kindly see attachment for the diagram

So in the diagram Displacement would be AC because it the shortest path between starting point and initial point.

So to calculate AC we can  Pythagorean theorem  because Δ ABC is right angled.

So phythagorus theorem is:-

H²=P²+B²    Where H=height ,P=perpendicular and B=Base

So in the given figure :-

Perpendicular=AB=AE+EB=3+3=6 KM

Base=BC=CF-BF=12-4=8 KM

Now we know base and height .So we can calculate the hypotenuse

AC²=8²+6²

AC²=64+36

AC²=100

AC=√100

AC=10 KM  

So AC=10 km which implies that the displacement is 10 kms.