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| <h3 id="back-link"><a href="../../#1.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.2">$\leftarrow$Back</a></h3> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.2.20.$ The body starts moving from point $A$ and moves first equidistant for time $t_0$, then with the same modulo acceleration — equidistant. After what time from the beginning of the movement, the body will return to point $A$? | | $1.2.20.$ The body starts moving from point $A$ and moves first equidistant for time $t_0$, then with the same modulo acceleration — equidistant. After what time from the beginning of the movement, the body will return to point $A$? |
| </p> | | </p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| During the time $t_0$ the velocity of the body became equal to $v_0 = at_0$, and the coordinate | | During the time $t_0$ the velocity of the body became equal to $v_0 = at_0$, and the coordinate |
| $$x_0 = \frac{at_0^2}{2}$$ | | $$x_0 = \frac{at_0^2}{2}$$ |
| It takes time $t_1$ for the body to travel the distance $x_1$: | | It takes time $t_1$ for the body to travel the distance $x_1$: |
| $$-x_1 = v_0 t - \frac{at^2}{2}$$ | | $$-x_1 = v_0 t - \frac{at^2}{2}$$ |
| Solving a quadratic equation: | | Solving a quadratic equation: |
| $$t_1^2 - 2t_0 t_1-t_0^2=0$$ | | $$t_1^2 - 2t_0 t_1-t_0^2=0$$ |
| We choose a positive root: | | We choose a positive root: |
| $$t_1 = (1+\sqrt{2})t_0$$ | | $$t_1 = (1+\sqrt{2})t_0$$ |
| Round trip time: | | Round trip time: |
| $$T = t_0+t_1=(2+\sqrt{2})t_0$$ | | $$T = t_0+t_1=(2+\sqrt{2})t_0$$ |
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| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$t =(2+\sqrt{2})t_0$$ | | $$t =(2+\sqrt{2})t_0$$ |
| </p> | | </p> |
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