Particle motion: position, velocity, acceleration

Problem 2.1617 · hard

1. \[ \frac{d}{d t} \left(2 t^{3} - 24 t^{2} + 90 t + 7\right) = 6 t^{2} - 48 t + 90 \]
   Velocity is the derivative of position.✓ Proved
2. \[ \frac{d}{d t} \left(6 t^{2} - 48 t + 90\right) = 12 t - 48 \]
   Acceleration is the derivative of velocity.✓ Proved
3. \[ 6 t^{2} - 48 t + 90 = \left(t - 3\right) \left(6 t - 30\right) \]
   v = 0 at t = 3 and t = 5: the particle is at rest.✓ Proved
4. v changes sign at each of those times, so the particle turns around there. Distance adds the length of each leg: [0, 3], [3, 5], [5, 6].
   Reviewed
5. \[ 8 + 8 + 108 = 124 \]
   |s(3) − s(0)| + |s(5) − s(3)| + |s(6) − s(5)|, with s = 7, 115, 107, 115.✓ Proved
6. For comparison, the displacement is s(6) − s(0) = 108 m.
   Reviewed