✅ 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 area 0→2 is 3.7 + 9.7 = 13.4 m
• Total Area 2→7 is 2(13.7) + 2(15.7) + 9.7(5.3) = 68.5 m
• This is close to 69 m from the Distance read‑offs

3) 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).

4) 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