A particle moves along a straight line such that the square of its velocity ( v^2 ) varies linearly with its position ( x ). If v^2 = 20 m ^2 s ⁻² at x = 5 m and v^2 = 100 m ^2 s ⁻² at x = 25 m , the magnitude of the acceleration of the par

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A particle moves along a straight line such that the square of its velocity ( v^2 ) varies linearly with its position ( x ). If v^2 = 20 m ^2 s ⁻² at x = 5 m and v^2 = 100 m ^2 s ⁻² at x = 25 m , the magnitude of the acceleration of the particle is _____ m s ⁻² .

Quantrex Academy · Accepted Answer

Correct answer: 2

Given that v^2 varies linearly with x , we can write the relation as: v^2 = mx + c The slope m of the v^2 versus x graph is: m = v₂^2 - v₁^2 x₂ - x₁ = 100 - 20 25 - 5 = 80 20 = 4 m s ⁻² Differentiating the equation v^2 = mx + c with respect to x , we get: 2v dv dx = m Since acceleration a = v dv dx , we have: 2a = m a = m 2 = 4 2 = 2 m s ⁻² Answer: 2

A particle moves along a straight line such that the square of its velocity ( v^2 ) varies linearly with its position ( x ). If v^2 = 20 m ^2 s ⁻² at x = 5 m and v^2 = 100 m ^2 s ⁻² at x = 25 m , the magnitude of the acceleration of the particle is _____ m s ⁻² .

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