✅ Adventure 6 — Solutions

✳ Functions

• a(t)=8-2t
• v(t)=t(8-t)=8t-t²
• s(t)=4t²-(1/3)t³

1) Distance read‑offs

• s(2)=13 m
• s(7)=82 m → distance 2→7 is 82-13=69 m

2) Area under v(t)

• Total distance 0→8 is the area under v(t).
• Exact integral: ∫₀⁸ (8t-t²)dt = 85.33 m

3) Rectangle estimates (0→8)

• Δt = 1 : 84.00 m
• Δt = 0.5 : 85.00 m
• Δt = 0.25 : 85.25 m
• These converge toward 85.33 m.

4) Velocity from acceleration area (0→8)

• Signed area under a(t) from 0→8 is zero (positive triangle cancels negative triangle).
• So v(8)=0 m/s (matches the velocity graph).

5) Acceleration check (0→5)

• From the velocity curve: Δv = v(5)-v(0)=15 m/s
• From acceleration area: area 0→4 is 16, area 4→5 is −1, net 15 m/s.

★ Calculus link

• s'(t)=v(t)
• v'(t)=a(t)
• Integrals reverse derivatives: ∫v(t)dt = s(t)+C