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Velocity and acceleration of a particle

Short answer

Differentiate: v = dr/dt = 2t i + 4 j + 4t k, a = dv/dt = 2 i + 4 k. At t=1, v = 2i+4j+4k, magnitude |v| = 6; a = 2i+4k, magnitude |a| = 2√5 ≈ 4.47.

The position vector of a moving particle at any time t is given by r = (t² + 1) i + (4t − 3) j + (2t² − 6) k. Find the velocity and acceleration at t = 1. Also find their magnitudes.

Velocity v = dr/dt and acceleration a = d²r/dt².

r = (t² + 1) i + (4t − 3) j + (2t² − 6) k
v = dr/dt = 2t i + 4 j + 4t k
a = dv/dt = 2 i + 0 j + 4 k

At t = 1:

v = 2 i + 4 j + 4 k
a = 2 i + 4 k
|v| = √(4 + 16 + 16) = √36 = 6
|a| = √(4 + 0 + 16) = √20 = 2√5 ≈ 4.47

Answer: v(1) = 2i + 4j + 4k with |v| = 6; a(1) = 2i + 4k with |a| = 2√5 ≈ 4.47.

✓ Answer checked by Claude Opus 5.5 (high effort)

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