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A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time \(\mathrm{t}\) is proportional to
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The correct answer is:
\(\mathrm{t}^{\frac{3}{2}}\)
\(\begin{aligned}
& \text {Hints: } \mathrm{P}=\mathrm{Fv}=\mathrm{m} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{v} \\
& \int \mathrm{vdv}=\int \mathrm{P} / \mathrm{mdt} ; \quad \frac{\mathrm{v}^2}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \\
& V=\sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{1}{2}} ; \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}} \mathrm{t}^{\frac{1}{2}} \\
& \int d x=\sqrt{\frac{2 p}{m}} \int t^{\frac{1}{2}} d t ; \quad x=\sqrt{\frac{2 p}{m}} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3} \sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{3}{2}} \\
& \mathrm{x} \alpha \mathrm{t}^{\frac{3}{2}} \\
&
\end{aligned}\)
& \text {Hints: } \mathrm{P}=\mathrm{Fv}=\mathrm{m} \cdot \frac{\mathrm{dv}}{\mathrm{dt}} \cdot \mathrm{v} \\
& \int \mathrm{vdv}=\int \mathrm{P} / \mathrm{mdt} ; \quad \frac{\mathrm{v}^2}{2}=\frac{\mathrm{Pt}}{\mathrm{m}} \\
& V=\sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{1}{2}} ; \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2 \mathrm{p}}{\mathrm{m}}} \mathrm{t}^{\frac{1}{2}} \\
& \int d x=\sqrt{\frac{2 p}{m}} \int t^{\frac{1}{2}} d t ; \quad x=\sqrt{\frac{2 p}{m}} \frac{t^{\frac{3}{2}}}{\frac{3}{2}}=\frac{2}{3} \sqrt{\frac{2 p}{m}} \mathrm{t}^{\frac{3}{2}} \\
& \mathrm{x} \alpha \mathrm{t}^{\frac{3}{2}} \\
&
\end{aligned}\)
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