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Acceleration proportional to time; distance travelled

Short answer

Acceleration a = –2t; integrating gives v = 16 – t², x = 4.33 + 16t – t³/3. At t=7 s: v = –33 m/s, x = 2.00 m. The particle reverses at t=4 s; total distance travelled = 42.67 + 45.00 = 87.67 m.

The acceleration of a particle is directly proportional to the time (t). At time (t) = 0, the velocity of the particle is v = 16 m/sec. Knowing that velocity (v) = 15 m/sec, position (x) = 20 m at time (t) = 1 sec, determine the velocity, the position and total distance travelled when time (t) = 7 sec.

Equations

a ∝ t, so a = kt.

  • v = v₀ + ∫kt dt = 16 + kt²/2
  • At t = 1 s, v = 15 m/s: 15 = 16 + k/2, so k = −2 m/s³ and a = −2t.
  • v = 16 − t² (m/s)
  • x = x₀ + 16t − t³/3. At t = 1 s, x = 20 m: 20 = x₀ + 16 − 1/3, so x₀ = 4.333 m.
  • x = 4.333 + 16t − t³/3 (m)

At t = 7 s

  • Velocity: v = 16 − 49 = −33 m/s (the particle moves in the negative direction).
  • Position: x = 4.333 + 112 − 343/3 = 4.333 + 112 − 114.333 = 2.00 m

Total distance travelled

The velocity is zero when 16 − t² = 0, so t = 4 s, and the direction of motion reverses.

  • At t = 4 s: x = 4.333 + 64 − 21.333 = 47.00 m
  • From t = 0 to 4 s: distance = 47.00 − 4.333 = 42.67 m
  • From t = 4 to 7 s: distance = 47.00 − 2.00 = 45.00 m

Total distance = 42.67 + 45.00 = 87.67 m (the displacement is only 2.00 − 4.33 = −2.33 m).

Answer: v = −33 m/s, x = 2.0 m, total distance = 87.67 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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