Physics, asked by rupeshkumarvemana60, 9 months ago

aA man has to go 50m due north,40m due east nad 20m due south to reach a field.(a)what is distance he has to walk to reach the field?.b what is his displacement from his house to the field

Answers

Answered by Hemu1432
1

Answer:

Distance :

Distance is the total path covered by the man.

Distance travelled by man=50m+40m+20m=110m

Displacement : 

Displacement is the shortest path between source [house] and destination [ field]= AB

See the figure :

Draw Perpendicular BE to AC.

BE=CD=40m

AE=AC-CE=50-20=30

NOW, DISPLACEMENT AB=√AE²+EB²

AB=√30²+40²=50m

Angle with horizontal is given by tanθ=30/40=3/4

θ=tan⁻¹[3/4] direction north  to east.

∴His displacement from his house to field is 50m , tan⁻¹[3/4] direction north  to east.

Attachments:
Answered by Anonymous
0

\Large{\underline{\underline{\bf{Solution :}}}}

We know that,

Distance = Total Path covered.

So,

Distance = 50 + 20 + 40

→Distance = 110 m

\rule{200}{2}

Now,

Fir, Displacement we will find the side of the triangle.

So,

50 m - 20 m

→30 m

_________________________

\sf{→Displacement = \sqrt{(30)^2 + (40)^2}} \\ \\ \sf{→Displacement = \sqrt{1600 + 900}} \\ \\ \sf{→Displacement = \sqrt{2500}} \\ \\ \sf{→Displacement = \pm 50 \: m}

As, Displacement can't be negative.

So, Displacement = 50 m

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