A particle moves in a steady velocity field described by v = -xî+2yĵ-zk̂, where x, y and z are the instantaneous coordinates of the particle. What is the magnitude of the particle's acceleration at the point (1, 2, 4)?
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Question

A particle moves in a steady velocity field described by v = -xî+2yĵ-zk̂, where x, y and z are the instantaneous coordinates of the particle. What is the magnitude of the particle's acceleration at the point (1, 2, 4)?

Options

G1 Physics · Accepted Answer

The velocity components are vₓ = -x, vᵧ = 2y, v(z) = -z. For the moving particle, dx ÷ dt = vₓ = -x, dy ÷ dt = vᵧ = 2y, dz ÷ dt = v(z) = -z. Differentiate the velocity: a = dv ÷ dt = -(dx ÷ dt)î + 2(dy ÷ dt)ĵ - (dz ÷ dt)k̂. Substituting the coordinate rates: a = -(-x)î + 2(2y)ĵ -(-z)k̂ = xî+4yĵ+zk̂. At the point (1, 2, 4), a = î+8ĵ+4k̂. Therefore, |a| = √(1²+8²+4²) = √81 = 9 m·s⁻². Correct option: B.