Правка разделов «Statement», «Answer»

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@@ -1,7 +1,8 @@
### Statement
−$14.1.8.$ [Insert the problem 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
@@ -12,4 +13,4 @@Solution
#### Answer
−[Insert a concise answer or boxed result]
+Relative to the airplane: $\frac{Kc}{2+K}$, Relative to the ground: $\frac{[K+(2+K)\beta]c}{(2+K)+K\beta}$