A particle is moving along the locus: y = k√x (x > 0) with a constant speed ‘v’. At t = 0, it is at the origin and about to enter the first quadrant of x-y axes. At some later time t > 0, vₓ = vᵧ. At this moment, [aᵧ - aₓ] =
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Question

A particle is moving along the locus: y = k√x (x > 0) with a constant speed ‘v’. At t = 0, it is at the origin and about to enter the first quadrant of x-y axes. At some later time t > 0, vₓ = vᵧ. At this moment, [aᵧ - aₓ] =

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G1 Physics · Accepted Answer

For y = k√x, dy/dx = k/(2√x). At the instant vₓ = vᵧ, dy/dx = 1, so x = k²/4. Also y'' = -k/(4x³ᐟ²) = -2/k² at this point. Constant speed gives vₓ = vᵧ = v/√2 and v⃗·a⃗ = 0, so aᵧ = -aₓ. Using aᵧ = y''vₓ² + (dy/dx)aₓ gives aᵧ - aₓ = y''v²/2 = -v²/k². Hence option C.

A particle is moving along the locus: y = k√x (x &gt; 0) with a constant speed ‘v’. At t = 0, it is at the origin and about to enter the first quadrant of x-y axes. At some later time t &gt; 0, vₓ = vᵧ. At this moment, [aᵧ - aₓ] =

A particle is moving along the locus: y = k√x (x > 0) with a constant speed ‘v’. At t = 0, it is at the origin and about to enter the first quadrant of x-y axes. At some later time t > 0, vₓ = vᵧ. At this moment, [aᵧ - aₓ] =