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The position co-ordinates of a particle movingin a 3-D coordinate system is given byx = a costy = a sintand z = atThe speed of the particle

Question

The position co-ordinates of a particle movingin a 3-D coordinate system is given byx = a costy = a sintand z = atThe speed of the particle

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Solution

To find the speed of the particle, we need to calculate the magnitude of its velocity vector. The velocity vector is the derivative of the position vector with respect to time.

Given the position coordinates: x = a cos(ωt) y = a sin(ωt) z = aωt

We can differentiate each coordinate with respect to time to find the velocity components:

vx = dx/dt = -aω sin(ωt) vy = dy/dt = aω cos(ωt) vz = dz/dt = aω

Now, we can calculate the magnitude of the velocity vector (speed) using the formula:

speed = √(vx^2 + vy^2 + vz^2)

Substituting the velocity components, we have:

speed = √((-aω sin(ωt))^2 + (aω cos(ωt))^2 + (aω)^2) = √(a^2ω^2 sin^2(ωt) + a^2ω^2 cos^2(ωt) + a^2ω^2) = √(a^2ω^2 (sin^2(ωt) + cos^2(ωt) + 1)) = √(a^2ω^2 (1 + 1)) = √(a^2ω^2 (2)) = √(2a^2ω^2) = √2aω

Therefore, the speed of the particle is √2aω.

This problem has been solved

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