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Velocity and acceleration of moving particle at t=1

Short answer

Velocity v = 2i + 4j + 4k, magnitude 6. Acceleration a = 2i + 0j + 4k, magnitude 2√5 ≈ 4.47.

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

r = (t² + 1) i + (4t − 3) j + (2t² − 6) k.

Velocity and acceleration

v = dr/dt = 2t i + 4 j + 4t k
f = d²r/dt² = 2 i + 0 j + 4 k

At t = 1

v = 2i + 4j + 4k
f = 2i + 4k

Magnitudes

|v| = √(4 + 16 + 16) = √36 = 6
|f| = √(4 + 0 + 16) = √20 = 2√5 ≈ 4.47

Answer: at t = 1, velocity = 2i + 4j + 4k with magnitude 6; acceleration = 2i + 4k with magnitude 2√5 ≈ 4.47.

✓ Answer checked by Claude Opus 5.5 (high effort)

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