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Position when velocity and acceleration given

Short answer

When v=2.5 m/s, solving t²-6t+8=2.5 gives t=1.129 s (x=20.69 m) and t=4.871 s (x=21.31 m). When a=3.6 m/s², 2t-6=3.6 gives t=4.8 s, x=21.14 m.

Motion of a particle is defined by a relation x = t³/3 − 3t² + 8t + 15. Determine the position of particle when velocity is 2.5 m/sec. Also determine the position of particle when acceleration is 3.6 m/sec².

Velocity and acceleration

  • x = t³/3 − 3t² + 8t + 15
  • v = dx/dt = t² − 6t + 8 (m/s)
  • a = dv/dt = 2t − 6 (m/s²)

Position when v = 2.5 m/s

t² − 6t + 8 = 2.5, so t² − 6t + 5.5 = 0, giving t = 3 ± √3.5 = 4.871 s or 1.129 s.

  • t = 4.871 s: x = 4.871³/3 − 3(4.871)² + 8(4.871) + 15 = 38.52 − 71.18 + 38.97 + 15 = 21.31 m
  • t = 1.129 s: x = 0.480 − 3.824 + 9.033 + 15 = 20.69 m

Position when a = 3.6 m/s²

2t − 6 = 3.6, so t = 4.8 s. x = 4.8³/3 − 3(4.8)² + 8(4.8) + 15 = 36.86 − 69.12 + 38.4 + 15 = 21.14 m (v at that instant = 4.8² − 6(4.8) + 8 = 2.24 m/s).

Answer: x = 21.31 m or 20.69 m when v = 2.5 m/s; x = 21.14 m when a = 3.6 m/s².

✓ Answer checked by Claude Opus 5.5 (high effort)

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