Câu hỏi 32

Xét tỉ số \(\dfrac{x}{v}\)

Ta có: \({\left( {\dfrac{x}{v}} \right)^\prime } = \dfrac{{x'.v - x.v'}}{{{v^2}}} = \dfrac{{{v^2} - x.a}}{{{v^2}}}\)

Chú ý: \(a =  - {\omega ^2}x \Rightarrow {\left( {\dfrac{x}{v}} \right)^\prime } = \dfrac{{{v^2} + {\omega ^2}{x^2}}}{{{v^2}}} = 1 + \dfrac{{{\omega ^2}{x^2}}}{{{v^2}}}\)

Công thức độc lập với thời gian:

\({\omega ^2}{x^2} + {v^2} = {\omega ^2}{A^2} \Rightarrow {v^2} = {\omega ^2}\left( {{A^2} - {x^2}} \right) \Rightarrow {\left( {\dfrac{x}{v}} \right)^\prime } = 1 + \dfrac{{{x^2}}}{{{A^2} - {x^2}}}\)

Theo đề bài ta có:

\(\begin{array}{l}\dfrac{{{x_1}}}{{{v_1}}} + \dfrac{{{x_2}}}{{{v_2}}} = \dfrac{{{x_3}}}{{{v_3}}} \Rightarrow {\left( {\dfrac{{{x_1}}}{{{v_1}}}} \right)^\prime } + {\left( {\dfrac{{{x_2}}}{{{v_2}}}} \right)^\prime } = {\left( {\dfrac{{{x_3}}}{{{v_3}}}} \right)^\prime }\\ \Rightarrow 1 + \dfrac{{{x_1}^2}}{{{A^2} - {x_1}^2}} + 1 + \dfrac{{{x_2}^2}}{{{A^2} - {x_2}^2}} = 1 + \dfrac{{{x_3}^2}}{{{A^2} - {x_3}^2}}\\ \Rightarrow 1 + \dfrac{{{4^2}}}{{{{10}^2} - {4^2}}} + 1 + \dfrac{{{3^2}}}{{{{10}^2} - {3^2}}} = 1 + \dfrac{{{x_3}^2}}{{{{10}^2} - {x_3}^2}} \Rightarrow {x_3} = 7,5\,\,\left( {cm} \right)\end{array}\)

Chọn D.