Question

A particle 20 metre in north east direction how much displacement it covers in eastward direction




Guys plz answer I need step by step explanation ​

Answer 1

Answer:

• Distance = 30 m + 40 m + 20 m.
• Displacement Vector = 30 j + 40 i + (-20) j = 40 i + 10 j.
• Magnitude of displacement =  \sqrt{40^2+10^2} =  10\sqrt{17}

Explanation:

Answer 2

The displacement it covers in East direction os 14.14 meters.

GIVEN

Distance travelled by a particle in North-East direction = 20 meter.

TO FIND

Displacement in the Eastward direction.

SOLUTION

We can simply solve the above problem as follows;

It is given that the particles moves in an North-East direction.

Let the distance covered by the particle in North-East direction be OA (d) = 20 meter.

This displacement vector has 2 component,

Vertical component = dsinθ

and

Horizontal component = dCosθ

Where, θ = 45°

This horizontal component is equal to the displacement in east direction.

Let the displacement in east direction be OB

$\cos(45) = \frac{OA}{OB}$

$\frac{1}{ \sqrt{2} } = \frac{OA}{20}$

$OA = \frac{20}{ \sqrt{2} }$

OA = 10 × 1.414

= 14.14

Hence, The displacement it covers in East direction os 14.14 meters.

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