甲乙丙相遇问题

Câu hỏi

甲、乙两人从相距490米的A、B两地同时步行出发，相向而行。丙与甲同时从A地出发，在甲、乙二人之间来回跑步（遇到乙立即返回，遇到甲也立即返回）。已知丙每分钟跑240米，甲每分钟走40米，当丙第一次折返回来并与甲相遇时，甲、乙二人相距210米，那么乙每分钟走多少米？甲下一次遇到丙时，甲、乙相距多少米？

Suy nghĩ

Analyzing the setup

I’m breaking down the initial positions and movements of persons A, B, and C, focusing on their speeds and the distance between them. Person C's continuous back-and-forth running pattern is relevant for further analysis.

Reassessing the meeting dynamics

I’m re-evaluating the problem to ensure clarity. C's meeting with A after the first turn-around occurs when A and B are 210 meters apart. This crucial detail affects their subsequent interactions.

Calculating speeds and distances

I'm calculating the speeds and distances for persons A, B, and C. At the initial turn-around, A and B are 210 meters apart. Next, I’ll determine their meeting dynamics and distances.

Figuring out meeting time

OK, let me see: B and C's meeting time is t1, calculated as 490 / (240 + vB), using their speeds and starting positions. Now, I'll determine when C meets A again.

Tracking C's journey

OK, let's think through it: C heads back towards A at 240 m/min after meeting B. Calculating C's position at time t2 involves considering both t1 and the distance covered in the return interval.

Piecing together the meeting

I’ve figured out that C's meeting time with A at t2 is 12/7 times t1. Now, I’m working out A and B’s distance at t2, which is 210 meters.

Calculating B's speed

I’m figuring out B's speed using the distance between A and B at t2. By substituting t2 and solving for B's speed, I piece together the necessary calculations.

Figuring out B's speed

I calculated B's speed to be 60 m/min by simplifying the equation and isolating terms. This helps understand the dynamics of A, B, and C's movements.

Charting C's path

After meeting A at t2, C heads towards B, aiming for t3. After t3, C returns to meet A again at t4, maintaining the cycle.

Tracking C's movements

Now we’re thinking through C’s path: after meeting A at t2, C heads towards B, then turns back to meet A again at each subsequent meeting, forming a continuous cycle.

Tracking C's movements

I verified that at t2, A and B are 210 meters apart, and C is 112 meters away from A. Now, I’ll calculate when C will meet B next.

Spotting the next gathering

I calculated that C and B will meet next at 3.5 minutes, at 280 meters from A, as C heads towards B after meeting A at t2.

Thought about walking and running motion，持续 23 秒

Trả lời

分析过程：

设甲、乙、丙三人同时开始运动，初始位置分别为：

甲在A点（记为坐标0处）
乙在B点（记为坐标490米处）
丙与甲同处A点（0米处）

已知条件：

丙的速度： $v_{丙} = 240$ 米/分
甲的速度： $v_{甲} = 40$ 米/分
乙的速度： $v_{乙}$ 米/分（未知，需要求出）

运动方向：甲从A向B方向前进，乙从B向A方向前进，丙在两者之间来回奔跑（与一端相遇立刻折返）。

第一阶段：丙首次由A向B跑并在某时刻与B相遇
设丙与B第一次相遇的时间为 $t_1$ 分钟。此时各自位置为：

甲的位置： $x_{甲}(t_1) = 0 + 40t_1$
乙的位置： $x_{乙}(t_1) = 490 - v_{乙}t_1$
丙的位置： $x_{丙}(t_1) = 0 + 240t_1$

丙与B相遇意味着丙、乙位置相同，即：

240t_1 = 490 - v_{乙}t_1

240t_1 + v_{乙}t_1 = 490

(240 + v_{乙})t_1 = 490

t_1 = \frac{490}{240 + v_{乙}}

第二阶段：丙在与B相遇后立即折返向A跑，并在时间 $t_2$ 时与甲相遇
丙在 $t_1$ 时的位置为 $240t_1$ 米处，此后丙以240米/分速度朝A方向跑。在时间 $t_2$ （> $t_1$ ）与甲相遇。

甲的位置： $x_{甲}(t_2) = 40t_2$
丙在 $t_1$ 时位置为 $240t_1$ ，此后向A方向跑，速度为240米/分。从 $t_1$ 到 $t_2$ 的时间间隔为 $t_2 - t_1$ 分钟，此期间丙返回路程为 $240(t_2 - t_1)$ 米，所以丙的位置为：

x_{丙}(t_2) = 240t_1 - 240(t_2 - t_1) = 240(2t_1 - t_2)

两者相遇时：

x_{甲}(t_2) = x_{丙}(t_2)

40t_2 = 240(2t_1 - t_2)

展开：

40t_2 = 480t_1 - 240t_2

40t_2 + 240t_2 = 480t_1

280t_2 = 480t_1

t_2 = \frac{480}{280}t_1 = \frac{12}{7}t_1

回顾已有关系：

t_1 = \frac{490}{240 + v_{乙}}, \quad t_2 = \frac{12}{7}t_1 = \frac{12}{7}\cdot\frac{490}{240 + v_{乙}}.

已知条件：当丙第二次与甲相遇(即时间 $t_2$ )时，甲、乙二人相距210米
此时二人的位置为：

x_{甲}(t_2) = 40t_2

x_{乙}(t_2) = 490 - v_{乙}t_2

两人之间的距离为：

x_{乙}(t_2) - x_{甲}(t_2) = [490 - v_{乙}t_2] - [40t_2] = 490 - (v_{乙} + 40)t_2

给定此距离为210米：

490 - (v_{乙} + 40)t_2 = 210

490 - 210 = (v_{乙} + 40)t_2

280 = (v_{乙} + 40)t_2

代入 $t_2 = \frac{12}{7}\cdot\frac{490}{240 + v_{乙}}$ :

280 = (v_{乙} + 40)\cdot\frac{12}{7}\cdot\frac{490}{240 + v_{乙}}

化简系数：

\frac{12}{7}\cdot 490 = 12 \cdot 70 = 840

故有：

280 = (v_{乙} + 40)\cdot\frac{840}{240 + v_{乙}}

两边乘以 $240 + v_{乙}$ :

280(240 + v_{乙}) = 840(v_{乙} + 40)

除以280：

240 + v_{乙} = 3(v_{乙} + 40)

展开右侧：

240 + v_{乙} = 3v_{乙} + 120

移项：

240 - 120 = 3v_{乙} - v_{乙}

120 = 2v_{乙}

v_{乙} = 60 \text{米/分}

至此，乙的速度求出：每分钟60米。

求解甲下一次与丙相遇时，甲、乙间距
目前已知：

$v_{甲} = 40$ 米/分
$v_{乙} = 60$ 米/分
$v_{丙} = 240$ 米/分

回顾时间点：

$t_1$ 为丙第一次与乙相遇时间
$t_2$ 为丙返回后第一次与甲相遇时间

我们已知 $t_1,t_2$ 数值可以代入求出：

v_{乙}=60,\quad t_1=\frac{490}{240+60}=\frac{490}{300}=1.6333\text{分}

t_2=\frac{12}{7}t_1=\frac{12}{7}\cdot1.6333= \frac{12}{7}\cdot\frac{490}{300}=2.8 \text{分}

在 $t_2=2.8$ 分时位置检查：

x_{甲}(2.8)=40\times2.8=112\text{米}

x_{乙}(2.8)=490-60\times2.8=490-168=322\text{米}

间距为210米，符合题意。

此时丙与甲在112米处相遇，丙再次转头向B跑。

丙下一次与B相遇：
设此时为 $t_3$ 分钟。
$t_2=2.8$ 分时丙在112米处转向B方向，以240米/分前进。

对 $t>2.8$ :

x_{丙}(t)=112 + 240(t-2.8)

丙与B相遇条件：

x_{丂}(t)=490-60t

设相遇时间为 $t_3$ :

112 + 240(t_3 - 2.8)=490 - 60t_3

112 + 240t_3 - 672 = 490 - 60t_3

240t_3 -560=490 -60t_3

240t_3 +60t_3=490+560

300t_3=1050

t_3=3.5\text{分}

相遇位置：

x_{丙}(3.5)=112+240(3.5-2.8)=112+240\times0.7=112+168=280\text{米}

x_{乙}(3.5)=490-60\times3.5=490-210=280\text{米}

验证无误。

此时丙在280米处与乙相遇并再次转向A。

丙再次与甲相遇的时间 $t_4$ ：
$t_3=3.5$ 分时丙在280米处转向A，以240米/分回跑。

对 $t>3.5$ :

x_{丙}(t)=280 -240(t-3.5)=280 -240t +840=1120-240t

甲的位置： $x_{甲}(t)=40t$

相遇条件：

40t=1120-240t

40t+240t=1120

280t=1120

t=4\text{分}

$t_4=4$ 分丙再次与甲相遇。

此时甲、乙位置：

x_{甲}(4)=40\times4=160\text{米}

x_{乙}(4)=490-60\times4=490-240=250\text{米}

两人相距：

250 -160 =90\text{米}

最终答案：
(1) 乙的速度为60米/分。
(2) 甲下一次（即在丙第二次折返后再次与甲相遇）与丙相遇时，甲、乙相距90米。

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