A particle moves in the x-y plane with velocity v = (3 t^2) i + (4 t) j in SI units. If the particle was at the origin (0, 0) at t = 0, what is the radius of curvature of its trajectory at t = 1 s?

Answer and explanation

Correct answer: C. 125 / 12 m

  1. Step 1: Velocity at t = 1: v = 3 i + 4 j. Speed v = sqrt(3^2 + 4^2) = 5 m/s.
  2. Step 2: Acceleration: a = dv/dt = 6t i + 4 j. At t = 1: a = 6 i + 4 j.
  3. Step 3: Cross product |v × a| = |3(4) - 4(6)| = |12 - 24| = 12 m^2/s^3.
  4. Step 4: Normal acceleration a_n = |v × a| / v = 12 / 5 m/s^2.
  5. Step 5: Radius of curvature R = v^2 / a_n = 5^2 / (12/5) = 125 / 12 m.

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