Question

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Answer 1

Answer :

The position vector of a particle r as a function of time is given by

• r = 4sin(2πt) i + 4cos(2πt) j

We have to find distance covered by particle in 4s$.$

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◈ Position of particle at t = 0s :

➝ r = 4sin(2π×0) i + 4cos(2π×0) j

➝ r = 4sin(0) i + 4cos(0) j

➝ r = 4(0) i + 4(1) j

➝ r = 4 j

◈ Position of particle at t = 1s :

➝ r₁ = 4sin(2π×1) i + 4cos(2π×1) j

➝ r₁ = 4sin(2π) i + 4cos(2π) j

➝ r₁ = 4(0) i + 4(1) j

➝ r₁ = 4 j

So we can say that, particle comes back to the initial position after 1s.

➢ x-component of r = 4sin(2πt) i

➢ y-component of r = 4cos(2πt) j

Equation of circle : R² = x² + y²

⟶ R² = [4sin(2πt)]² + [4cos(2πt)]²

⟶ R² = 4²[sin²(2πt) + cos²(2πt)]

⟶ R² = 4²(1)

⟶ R = 4 m

Hence radius of the circle is 4m.

We know that,

Perimeter of circle = 2πR

Therefore distance covered by particle in 1 second will be equal to 2πR because particle comes back to the initial position after 1 second.

◈ Distance covered in 4 seconds :

⇒ d = 4 × 2πR

⇒ d = 4 × 2π(4)

⇒ d = 4 × 8π

⇒ d = 32π m

Nice Question! :)