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| <title>The car moves with speed v away from a long wall, moving at an angle lpha to it. At the moment when the distance to the wall equals l, the driver gives a short beep. How far will the car travel before the chauffeur hears the echo? The speed of sound in the air is c.</title> | | <title>The car moves with speed v away from a long wall, moving at an angle lpha to it. At the moment when the distance to the wall equals l, the driver gives a short beep. How far will the car travel before the chauffeur hears the echo? The speed of sound in the air is c.</title> |
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| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.1.18^*.$ The car moves with speed $v$ away from a long wall, moving at an angle $\alpha$ to it. At the moment when the distance to the wall equals $l$, the driver gives a short beep. How far will the car travel before the chauffeur hears the echo? The speed of sound in the air is $c$. | | $1.1.18^*.$ The car moves with speed $v$ away from a long wall, moving at an angle $\alpha$ to it. At the moment when the distance to the wall equals $l$, the driver gives a short beep. How far will the car travel before the chauffeur hears the echo? The speed of sound in the air is $c$. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="180" /> | | loading="lazy" width="180" /> |
| <figcaption> | | <figcaption> |
| For problem $1.1.18^*$ | | For problem $1.1.18^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
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| <p> | | <p> |
| Since the sound signal from the car is immediately reflected from the wall, it is possible to mirror the signal from the car, as ‘if’ the sound signal catches up with the imaginary image of a moving car | | Since the sound signal from the car is immediately reflected from the wall, it is possible to mirror the signal from the car, as ‘if’ the sound signal catches up with the imaginary image of a moving car |
| </p> | | </p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="Mirror.png" alt="1.1.18" | | <img src="Mirror.png" alt="1.1.18" |
| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| Mirror reflection with respect to the wall | | Mirror reflection with respect to the wall |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Further, we take the fact that in time $t$, the sound travels distance $ct$ and the car distance $vt$ along the line of its motion: | | Further, we take the fact that in time $t$, the sound travels distance $ct$ and the car distance $vt$ along the line of its motion: |
| </p> | | </p> |
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| <center> | | <center> |
| <figure> | | <figure> |
| <img src="sol.jpg" alt="1.1.18" | | <img src="sol.jpg" alt="1.1.18" |
| loading="lazy" width="400" /> | | loading="lazy" width="400" /> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Let us write the Pythagorean Theorem for the resulting right triangle: | | Let us write the Pythagorean Theorem for the resulting right triangle: |
| </p> | | </p> |
| | | |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$(vt\,cos\;\alpha)^2+(vt\,sin\;\alpha + 2l)^2 = c^2t^2$$ | | $$(vt\,cos\;\alpha)^2+(vt\,sin\;\alpha + 2l)^2 = c^2t^2$$ |
| </p> | | </p> |
| <p> | | <p> |
| Rewrite the quadratic equation as: | | Rewrite the quadratic equation as: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$(v^2-c^2)t^2 + 4lvt\,sin\;\alpha + 4l^2= 0$$ | | $$(v^2-c^2)t^2 + 4lvt\,sin\;\alpha + 4l^2= 0$$ |
| </p> | | </p> |
| <p> | | <p> |
| We get a positive root: | | We get a positive root: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$$ | | $$t=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$$ |
| </p> | | </p> |
| <p> | | <p> |
| Since the distance travelled by the car until the driver hears the sound is $x=vt$, the distance is found as: | | Since the distance travelled by the car until the driver hears the sound is $x=vt$, the distance is found as: |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$\fbox{$t=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$}$$ | | $$\fbox{$t=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$}$$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$x=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$$ | | $$x=2l\frac{v\;sin\;\alpha+\sqrt{c^2-v^2\;cos^2\;\alpha}}{c^2-v^2}$$ |
| </p> | | </p> |
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