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Since the radius is constant, the derivative of the radius, R is zero and

(4) v⃗_B = θ̇Rê_θ + ⎛⎝ − (Pθ̇)/(2π)⎞⎠k̂

The acceleration of the helicopter can be found from eqn. 2↑ to be

a⃗_B = (R̈ − Rθ̇²ê_r + Rθ̈ê_θ + ⎛⎝ − (Pθ̈)/(2π)⎞⎠k̂

Again, since R is a constant, its derivatives are zero and the equation simplifies to

(5) a⃗_B = − Rθ̇²ê_r + Rθ̈ê_θ + ⎛⎝ − (pθ̈)/(2π)⎞⎠k̂

At this point, the values can be substituted into the equation. Given R = 0.3m and p = 0.2m, and θ̇ = .6t(r)/(s), then the v⃗_B and a⃗_B at t = 10s can be found. First, the second derivative of θ is found to be

θ̈ = 0.6(r)/(s²)

Substituting into the velocity equation for the helicopter gives

v⃗_B = 0.6(10)(0.3)ê_θ − ((0.2)(0.6)(10))/(2π)k̂ v⃗_B = 1.8ê_θ − (0.6)/(π)k̂ (m)/(s)

This velocity makes sense physically since the helicopter is moving around and down, not outward or inward. The acceleration can likewise be found as

a⃗_B = − (0.3)(.6(10))²ê_r + (0.3)(0.6)ê_θ − ((.2)(.6))/(2π)k̂ = − (0.3)(36)ê_r + 0.18ê_θ − (0.06)/(π)k̂ a⃗_B = − 10.8ê_r + 0.18ê_θ − (0.06)/(π)k̂ (m)/(s²)

Notice that the answers are given in cylindrical Coordinates (ê_r, ê_θ, k̂) not in Cartesian (î, ĵ, k̂)