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    <h1>✅ Adventure 6 — Solutions</h1>
    <div class="subtitle">
      Connect graph reading, rectangle estimates, and the calculus meaning of <b>area</b>.
    </div>

    <div class="two-col">
      <div class="left-col">

        <h2>✳ Functions</h2>
        <ul>
          <li><code>a(t)=8-2t</code></li>
          <li><code>v(t)=t(8-t)=8t-t²</code></li>
          <li><code>s(t)=4t²-(1/3)t³</code></li>
        </ul>

        <h2>1) Distance read‑offs</h2>
        <ul>
          <li><code>s(2)=13</code> m</li>
          <li><code>s(7)=82</code> m → distance 2→7 is <code>82-13=69</code> m</li>
        </ul>

        <h2>2) Area under v(t)</h2>
        <ul>
          <li>Total area 0→2 is 3.7 + 9.7 = 13.4 m</li>
          <li>Total Area 2→7 is 2(13.7) + 2(15.7) + 9.7(5.3) = 68.5 m</li>
          <li>This is close to 69 m from the Distance read‑offs</li>
        </ul>

        
        <h2>3) Velocity from acceleration area (0→8)</h2>
        <ul>
          <li>Signed area under <code>a(t)</code> from 0→8 is zero (positive triangle cancels negative triangle).</li>
          <li>So <code>v(8)=0</code> m/s (matches the velocity graph).</li>
        </ul>

        <h2>4) Acceleration check (0→5)</h2>
        <ul>
          <li>From the velocity curve: <code>Δv = v(5)-v(0)=15</code> m/s</li>
          <li>From acceleration area: area 0→4 is 16, area 4→5 is −1, net 15 m/s.</li>
        </ul>

        <h2>★ Calculus link</h2>
        <ul>
          <li><code>s'(t)=v(t)</code></li>
          <li><code>v'(t)=a(t)</code></li>
          <li>Integrals reverse derivatives: <code>∫v(t)dt = s(t)+C</code></li>
        </ul>

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Anon7 - 2021