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Find the velocity v(t) and speed ∥v(t)∥ of a particle whose motion is described by x=8t^3−36t^2, y=3t^2−18t+27, z=2

User Diatoid
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Final answer:

The velocity v(t) and speed ∥v(t)∥ of the particle can be found by differentiating the position vector and calculating its magnitude.

Step-by-step explanation:

To find the velocity v(t) and speed ∥v(t)∥ of the particle, we need to differentiate the given position vector with respect to time.

Given position vector: x = 8t^3 - 36t^2, y = 3t^2 - 18t + 27, z = 2

Velocity vector: v(t) = (dx/dt)i + (dy/dt)j + (dz/dt)k

By differentiating each component of the position vector, we get:
dx/dt = 24t^2 - 72t
dy/dt = 6t - 18
dz/dt = 0

Therefore, the velocity vector is given by:
v(t) = (24t^2 - 72t)i + (6t - 18)j

To find the speed, we calculate the magnitude of the velocity vector:
∥v(t)∥ = √((24t^2 - 72t)^2 + (6t - 18)^2)

Learn more about Velocity and Speed

User MarcM
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