Question

A particle moves through a distance of 8 m due East and then 6 m due North. Then the magnitude of
displacement is
WI
a) 2 m
b) 14 m
c) 6 m
d) 10 m


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Answer 1

Given :-

Distance travelled in the first case = 8 m east

Distance travelled in the second case = 6 m north

To Find :-

The magnitude of  displacement.

Solution :-

We know that,

• a = Side of right triangle
• b = Side of right triangle
• c = Hypotenuse

Displacement is the shortest path travelled by the body.

Using Pythagoras theorem,

$\underline{\boxed{\sf Pythagoras \ theorem=AC^2=AB^2+BC^2}}$

Given that,

BC = 6 cm

AB = 8 cm

Substituting their values,

AC² = (8)² + (6)²

AC² = 64 + 36

AC² = 100

By transposing,

AC = √100

AC = 10 m

Therefore, the magnitude of  displacement is 10 m.

Answer 2

Given:-

• A particle moves through a distance of 8 m East and then 6 m North.

To Find:-

• magnitude of displacement.

Formula Used:-

Pythagoras Theorem

a²+ b² = c²

• a = side of right triangle
• b = side of right triangle
• c = hypotenuse

Solution:-

Applying Pythagoras theorem in ∆ ABC

$AC {}^{2} = AB {}^{2} + BC {}^{2} \\ AC {}^{2} = {8}^{2} + {6}^{2} \\ AC {}^{2} = 64 + 36 \\ AC {}^{2} =100 \\ AC =10$

:. So displacement is 10 m