Новое решение
en/14.1.8.md
+15 −0
| @@ -0,0 +1,15 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $14.1.8.$ [Insert the problem statement] | ||
| + | |||
| + | ### 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 | ||
| + | |||
| + | \[\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},\] | ||
| + | |||
| + | which becomes $\beta c$ if $K=0$, i.e. intervals between reflected pulses equal intervals between emitted pulses. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,15 @@ | |||
| ### Statement | |||
| $14.1.8.$ [Insert the problem statement] | |||
| ### 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 | |||
| \[\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},\] | |||
| which becomes $\beta c$ if $K=0$, i.e. intervals between reflected pulses equal intervals between emitted pulses. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||