Question

A man has to go 50 metre due north, 40 metre due east and 20 metre due south to reach a field what is his displacement from his house to the field ???

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Answer 1

Answer:

Man goes 50 m to North , then 40 m to East and 20 due South. We need to find out the displacement of the whole journey.

First of all , we need to define Displacement :

Displacement is the shortest length between the starting and stopping point. It's a vector having both directions and magnitude.

Look.at the attached diagram to understand his trajectory from home to field.

So the diagram shows a Trapezium like figure which is formed in his journey . We need to find x.

Thos can be done using Pythagoras theorem:

$\bigstar \: \: {30}^{2} + {40}^{2} = {x}^{2}$

$= > {x}^{2} = 1600 + 900$

$= > {x}^{2} = 2500$

$= > x = 50 \: m$

So displacement is 50 m

Answer 2

Given:-

$\texttt{A man has to go 50 metre due north, 40 metre due east }\\\texttt{and 20 metre due south to reach a field.}$

To find:-

$\texttt{The displacement from his house to the field.}$

Solution:-

$\texttt{We know that displacement is the shortest length }\\\texttt{between the initial point and the final point.}$

♦ $\texttt{Distance travelled towards North is AB.}$

♦ $\texttt{Distance travelled towards East is BC.}$

♦ $\texttt{Distance travelled towards South is CD.}$

$\texttt{Displacement = AD}$

$\texttt{Let us draw a line segment joining D with BA at F.}$

• $\texttt{DF = BC, as BCDF forms a rectangle.}$

• $\texttt{Then CD = BF = 20 m.}$

$\texttt{Now, in triangle DAF, AF = 50 m - 20 m = 30 m, DF = 40 m and }\\\texttt{AD = x m(to be found).}$

$\texttt{By pythagoras theorem,}\\\\\tt{(AF)^2 + (DF)^2 = (AD)^2}\\\\\tt{\implies (30)^2+(40)^2=(x)^2}$

$\tt{\implies 900+1600=(x)^2}\\\\\tt{\implies x=\sqrt{2500} }\\\\\tt{\boxed{\implies x=AD=50\ m}}$

∴$\bold{\underline{So,\ the\ displacement\ of\ the\ man\ is\ 50\ m\ from\ his\ house\ to\ the\ field.}}$