Question

A man walks 50 m towards East, then 40 m towards North and then 20 m towards West. The magnitude of his displacement is

• A. 40 m
• B. 50 m
• C. 30 m
• D. 110 m

Answer 1

The correct option is B 50 m

The path followed by the man is represented in the given figure.
DE AB
AE = 30 m
DE = CB = 40 mAccording to the figure:
The shortest distance between the initial point A and the final point D is displacement AD.
In ΔADE, using Pythagoras Theorem:
$(AD{)}^2=(AE{)}^2+(DE{)}^2$
$\Rightarrow (AD{)}^2=(30{)}^2+(40{)}^2$
$\Rightarrow (AD{)}^2=900+1600$
$\Rightarrow AD=\sqrt{2500}m$
$\Rightarrow AD=50m$
Hence, the correct answer is option (b).