Hướng dẫn giải Thảo luận 6 (Trang 44 SGK Vật lý 10, Bộ Chân trời sáng tạo)

Rút ra phương trình liên hệ giữa gia tốc, vận tốc và độ dịch chuyển (7.6). Lời giải: Ta có: v = v0 + at => t = <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>t</mi><mo>=</mo><mfrac><mrow><mi>v</mi><mo>-</mo><msub><mi>v</mi><mn>0</mn></msub></mrow><mi>a</mi></mfrac></math> Thay t vào phương trình: <span id="MathJax-Element-2-Frame" class="mjx-chtml MathJax_CHTML" style="box-sizing: border-box; display: inline-block; line-height: 0; text-indent: 0px; text-align: left; text-transform: none; font-style: normal; font-weight: normal; font-size: 21.78px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; position: relative;" tabindex="0" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><msub><mi>v</mi><mn>0</mn></msub><mi>t</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>a</mi><msup><mi>t</mi><mn>2</mn></msup></math>"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><msub><mi>v</mi><mi>o</mi></msub><mi>t</mi><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>a</mi><msup><mi>t</mi><mn>2</mn></msup></math><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mn></mn></msup></math>  ta được:<span id="MathJax-Element-3-Frame" class="mjx-full-width mjx-chtml MathJax_CHTML" style="box-sizing: border-box; display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 21.78px; letter-spacing: normal; overflow-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 11.334em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;" tabindex="0" role="presentation" data-mathml="<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><msub><mi>v</mi><mn>0</mn></msub><mfenced><mfrac><mrow><mi>v</mi><mo>&#x2212;</mo><msub><mi>v</mi><mn>0</mn></msub></mrow><mi>a</mi></mfrac></mfenced><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>a</mi><msup><mfenced><mfrac><mrow><mi>v</mi><mo>&#x2212;</mo><msub><mi>v</mi><mn>0</mn></msub></mrow><mi>a</mi></mfrac></mfenced><mn>2</mn></msup><mspace linebreak="newline" /><mo>=</mo><mfrac><mrow><msup><mi>v</mi><mn>2</mn></msup><mo>&#x2212;</mo><msubsup><mi>v</mi><mn>0</mn><mn>2</mn></msubsup></mrow><mrow><mn>2</mn><mi>a</mi></mrow></mfrac><mo>&#x21D2;</mo><msup><mi>v</mi><mn>2</mn></msup><mo>&#x2212;</mo><msubsup><mi>v</mi><mn>0</mn><mn>2</mn></msubsup><mo>=</mo><mn>2</mn><mi>a</mi><mi>d</mi></math>"> <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>d</mi><mo>=</mo><msub><mi>v</mi><mi>o</mi></msub><mfenced><mfrac><mrow><mi>v</mi><mo>-</mo><msub><mi>v</mi><mn>0</mn></msub></mrow><mi>a</mi></mfrac></mfenced><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mi>a</mi><msup><mfenced><mfrac><mrow><mi>v</mi><mo>-</mo><msub><mi>v</mi><mn>0</mn></msub></mrow><mi>a</mi></mfrac></mfenced><mn>2</mn></msup><mspace linebreak="newline"/><mo>=</mo><mfrac><mrow><msup><mi>v</mi><mn>2</mn></msup><mo>-</mo><msubsup><mi>v</mi><mn>0</mn><mn>2</mn></msubsup></mrow><mrow><mn>2</mn><mi>a</mi></mrow></mfrac><mo>⇒</mo><msup><mi>v</mi><mn>2</mn></msup><mo>-</mo><msubsup><mi>v</mi><mn>0</mn><mn>2</mn></msubsup><mo>=</mo><mn>2</mn><mi>a</mi><mi>d</mi></math>

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Lý thuyết Gia tốc - Chuyển động thẳng biến đổi đều