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Normal, tangential acceleration; radius of curvature

Short answer

At t=4 s, v = 1294.54 m/s, a = 964.81 m/s². Tangential acceleration a_t = a·v/v = 963.6 m/s². Normal acceleration a_n = √(a²−a_t²) = 48.9 m/s². Radius of curvature ρ = v²/a_n = 3.43×10⁴ m.

The motion of a particle is defined by the position vector r = 3t²i + 4t³j + 5t⁴k where r is in meter and t is in second. Find the normal and tangential component of acceleration and the principal radius of curvature at the instant when t = 4 secs.

Velocity and acceleration vectors

r = 3t² i + 4t³ j + 5t⁴ k

  • v = dr/dt = 6t i + 12t² j + 20t³ k
  • a = dv/dt = 6 i + 24t j + 60t² k

At t = 4 s:

  • v = 24 i + 192 j + 1280 k (m/s), so v = √(24² + 192² + 1280²) = 1294.54 m/s
  • a = 6 i + 96 j + 960 k (m/s²), so a = √(6² + 96² + 960²) = 964.81 m/s²

Tangential component

a_t = a·v/v = (6 × 24 + 96 × 192 + 960 × 1280)/1294.54 = (144 + 18432 + 1228800)/1294.54 = 963.57 m/s²

Normal component

a_n = √(a² − a_t²) = √(964.81² − 963.57²) = 48.93 m/s²

Radius of curvature

ρ = v²/a_n = 1294.54²/48.93 = 34 250 m ≈ 34.25 km

Answer: a_t = 963.6 m/s², a_n = 48.9 m/s², ρ ≈ 3.43 × 10⁴ m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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