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Velocity and acceleration for r = 4θ², θ = t²/2

Short answer

At θ=80°, compute θ̇=1.671 rad/s, r=7.798 m, ṙ=18.67 m/s. Velocity: v_r=18.67 m/s, v_θ=13.03 m/s, magnitude 22.8 m/s. Acceleration: a_r=11.73 m/s², a_θ=70.19 m/s², magnitude 71.2 m/s².

A particle moves along a curved path defined by r = 4θ² and θ = t²/2 where r is in meters and t in seconds. Determine the velocity and acceleration of the particle when θ = 80°.

Polar components

Given r = 4θ² and θ = t²/2. At θ = 80° = 1.3963 rad:

  • θ̇ = t = √(2θ) = √2.7925 = 1.6711 rad/s, θ̈ = 1 rad/s²
  • r = 4θ² = 4(1.3963)² = 7.798 m
  • ṙ = 8θθ̇ = 8(1.3963)(1.6711) = 18.67 m/s
  • r̈ = 8θ̇² + 8θθ̈ = 8(2.7925) + 8(1.3963) = 33.51 m/s²

Velocity

  • v_r = ṙ = 18.67 m/s
  • v_θ = rθ̇ = 7.798 × 1.6711 = 13.03 m/s
  • v = √(18.67² + 13.03²) = 22.77 m/s, at an angle tan⁻¹(13.03/18.67) = 34.9° from the radial line.

Acceleration

  • a_r = r̈ − rθ̇² = 33.51 − 7.798(2.7925) = 33.51 − 21.78 = 11.73 m/s²
  • a_θ = rθ̈ + 2ṙθ̇ = 7.798(1) + 2(18.67)(1.6711) = 7.80 + 62.40 = 70.19 m/s²
  • a = √(11.73² + 70.19²) = 71.16 m/s², at 80.5° from the radial line.

Answer: v = 22.8 m/s (v_r = 18.67 m/s, v_θ = 13.03 m/s); a = 71.2 m/s² (a_r = 11.73 m/s², a_θ = 70.19 m/s²).

✓ Answer checked by Claude Opus 5.5 (high effort)

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