Правка раздела «Solution»
en/14.1.8.md
+1 −1
| ### Statement | |||
| $14.1.8.$ The plane and the rocket move in the same straight line and in the same direction. Aircraft speed $\beta c$. Light pulses are emitted from the aircraft at regular intervals, which, reflected from the rocket, arrive at the aircraft at intervals $K$ times longer than the intervals of the emitted pulses. Determine the speed of the rocket relative to the plane from observations from the plane and from observations from the Ground. | |||
| ### Solution | |||
| Let intervals between emitted pulses (in the frame of the airplane) be $T$. We interpret the phrase "intervals $K$ times longer than the intervals of the emitted pulses" for the reflected pulses received at the airplane to be $T+KT$. Suppose the airplane and the rocket coincide at time $t=0$ when the first pulse is emitted (and immediately reflected and received). When the second pulse is emitted at time $t=T$, the rocket is $vT$ ahead of the airplane, where $v$ is the speed of the rocket relative to the airplane. Thus, the second pulse will be reflected at the rocket at time $t=T+vT/(c-v)$ and received at the airplane at time $t=T+2vT/(c-v)$. Setting $KT=2vT/(c-v)$, we obtain $v=Kc/(2+K)$. Since the speed of the ground relative to the airplane is $-\beta c$, the speed of the rocket relative to the ground is | |||
| @@ -9,8 +9,8 @@Solution | |||
| \[\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},\] | |||
| − | which becomes $\beta c$ if $K=0$ | ||
| + | which becomes $\beta c$ if $K=0$; that is, if intervals between reflected pulses received at the airplane equal intervals between emitted pulses, then the airplane and the rocket have the same speed. | ||
| #### Answer | |||
| Relative to the airplane: $\frac{Kc}{2+K}$, Relative to the ground: $\frac{[K+(2+K)\beta]c}{(2+K)+K\beta}$ | |||
| ### Statement | ### Statement | ||
| $14.1.8.$ The plane and the rocket move in the same straight line and in the same direction. Aircraft speed $\beta c$. Light pulses are emitted from the aircraft at regular intervals, which, reflected from the rocket, arrive at the aircraft at intervals $K$ times longer than the intervals of the emitted pulses. Determine the speed of the rocket relative to the plane from observations from the plane and from observations from the Ground. | $14.1.8.$ The plane and the rocket move in the same straight line and in the same direction. Aircraft speed $\beta c$. Light pulses are emitted from the aircraft at regular intervals, which, reflected from the rocket, arrive at the aircraft at intervals $K$ times longer than the intervals of the emitted pulses. Determine the speed of the rocket relative to the plane from observations from the plane and from observations from the Ground. | ||
| ### Solution | ### Solution | ||
| Let intervals between emitted pulses (in the frame of the airplane) be $T$. We interpret the phrase "intervals $K$ times longer than the intervals of the emitted pulses" for the reflected pulses received at the airplane to be $T+KT$. Suppose the airplane and the rocket coincide at time $t=0$ when the first pulse is emitted (and immediately reflected and received). When the second pulse is emitted at time $t=T$, the rocket is $vT$ ahead of the airplane, where $v$ is the speed of the rocket relative to the airplane. Thus, the second pulse will be reflected at the rocket at time $t=T+vT/(c-v)$ and received at the airplane at time $t=T+2vT/(c-v)$. Setting $KT=2vT/(c-v)$, we obtain $v=Kc/(2+K)$. Since the speed of the ground relative to the airplane is $-\beta c$, the speed of the rocket relative to the ground is | Let intervals between emitted pulses (in the frame of the airplane) be $T$. We interpret the phrase "intervals $K$ times longer than the intervals of the emitted pulses" for the reflected pulses received at the airplane to be $T+KT$. Suppose the airplane and the rocket coincide at time $t=0$ when the first pulse is emitted (and immediately reflected and received). When the second pulse is emitted at time $t=T$, the rocket is $vT$ ahead of the airplane, where $v$ is the speed of the rocket relative to the airplane. Thus, the second pulse will be reflected at the rocket at time $t=T+vT/(c-v)$ and received at the airplane at time $t=T+2vT/(c-v)$. Setting $KT=2vT/(c-v)$, we obtain $v=Kc/(2+K)$. Since the speed of the ground relative to the airplane is $-\beta c$, the speed of the rocket relative to the ground is | ||
| @@ -9,8 +9,8 @@Solution | |||
| \[\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},\] | \[\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},\] | ||
| which becomes $\beta c$ if $K=0$ |
which becomes $\beta c$ if $K=0$; that is, if intervals between reflected pulses received at the airplane equal intervals between emitted pulses, then the airplane and the rocket have the same speed. | ||
| #### Answer | #### Answer | ||
| Relative to the airplane: $\frac{Kc}{2+K}$, Relative to the ground: $\frac{[K+(2+K)\beta]c}{(2+K)+K\beta}$ | Relative to the airplane: $\frac{Kc}{2+K}$, Relative to the ground: $\frac{[K+(2+K)\beta]c}{(2+K)+K\beta}$ | ||