Respuesta :
The expression for velocity is [tex]v(t)=\frac{\sqrt{2g}}{\sqrt[3]{x(t)}}R_{E}[/tex]
The expression for acceleration is [tex]a(t)=-\left ( \frac{gR_{E}^{2}}{[R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t]^\frac{4}{3}} \right )[/tex]
The expression for the rocket's position as measured from the center of the earth is
[tex]y(t)=[R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t]^\frac{2}{3}[/tex]
Let
[tex]x(t)=R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t[/tex]
We have,
[tex]y(t)=[x(t)]^\frac{2}{3}[/tex]
The velocity of the rocket as a function of time is given by
[tex]v(t)=\frac{\mathrm{d} y(t)}{\mathrm{d} t}=\frac{2}{3}[x(t)]^{-\frac{1}{3}}\frac{\mathrm{d} x(t)}{\mathrm{d} t}[/tex]
[tex]v(t)=\frac{\mathrm{d} y(t)}{\mathrm{d} t}=\frac{2}{3}\frac{1}{\sqrt[3]{x(t)}}\frac{\mathrm{d} x(t)}{\mathrm{d} t}[/tex]
Since
[tex]x(t)=R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t[/tex]
We have,
[tex]\frac{\mathrm{d} x(t)}{\mathrm{d} t}=3\sqrt{\frac{g}{2}}R_{E}[/tex]
[tex]v(t)=\frac{2}{3}\frac{1}{\sqrt[3]{x(t)}}\frac{\mathrm{d} x(t)}{\mathrm{d} t}[/tex]
[tex]v(t)=\frac{2}{3}\frac{1}{\sqrt[3]{x(t)}} X 3 \sqrt{\frac{g}{2}}R_{E}=\frac{\sqrt{2g}}{\sqrt[3]{x(t)}}R_{E}[/tex]
[tex]v(t)=\frac{\sqrt{2g}}{\left ( R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t \right )^\frac{1}{3}}R_{E}[/tex]
[tex]v(t)=\frac{\sqrt{2g}}{\sqrt[3]{x(t)}}R_{E}[/tex] .......(i)
The acceleration as a function of time is given by
[tex]a(t)=\frac{\mathrm{d} v(t)}{\mathrm{d} t}[/tex]
[tex]a(t)=\frac{\mathrm{d}}{\mathrm{d} t}\left (\frac{\sqrt{2g}}{\sqrt[3]{x(t)}}R_{E}\right )[/tex]
[tex]a(t)=\frac{\mathrm{d}}{\mathrm{d} t}\left ({\sqrt{2g}}R_{E}[x(t)]^{-\frac{1}{3}}\right )[/tex]
[tex]a(t)=\sqrt{2g}R_{E}\frac{\mathrm{d}}{\mathrm{d} t}[(x(t)]^{-\frac{1}{3}}[/tex]
[tex]a(t)=-\frac{1}{3}\sqrt{2g}R_{E} [(x(t)]^{-\frac{4}{3}}\frac{\mathrm{d} x(t)}{\mathrm{d} t}[/tex]
[tex]a(t)=-\frac{1}{3}\sqrt{2g}R_{E} [(x(t)]^{-\frac{4}{3}}3\sqrt{\frac{g}{2}}R_{E}[/tex]
[tex]a(t)=-gR_{E}^2 [(x(t)]^{-\frac{4}{3}}[/tex]
[tex]a(t)=-\left ( \frac{gR_{E}^{2}}{[x(t)]^\frac{4}{3}} \right )[/tex]
[tex]a(t)=-\left ( \frac{gR_{E}^{2}}{[R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t]^\frac{4}{3}} \right )[/tex] .......(ii)
Therefore, the expression for velocity is [tex]v(t)=\frac{\sqrt{2g}}{\sqrt[3]{x(t)}}R_{E}[/tex]
& the expression for acceleration is [tex]a(t)=-\left ( \frac{gR_{E}^{2}}{[R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t]^\frac{4}{3}} \right )[/tex]
Read more about Velocity and Accelaration:
https://brainly.com/question/1597434
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In Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A better expression for a rocket's position measured from the center of the Earth is given by [tex]y(t)=[R_{E}^{\frac{3}{2}}+3\sqrt{\frac{g}{2}}R_{E}t]^\frac{2}{3}[/tex]
where RE is the radius of the Earth (6.38 ✕ 106 m) and g is the constant acceleration of an object in free fall near the Earth's surface (9.81 m/s2).
(a) Derive expressions for v(t) and a(t).
(Use the following as necessary: g, RE, and t. Do not substitute numerical values; use variables only.)
