1997 PMWC Problems/Problem T4

Problem

In one morning, a ferry traveled from Hong Kong to Kowloon and another ferry traveled from Kowloon to Hong Kong at a different speed. They started at the same time and met first time at $8:20$ . The two ferries then sailed to their destinations, stopped for $15$ minutes and returned. The two ferries met again at $9:11$ . Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey?

Solution

Time $t$ , velocity $v$ , distance $a$ , formula $v=\frac{a}{t}$ or $a=v \cdot t$ . Time of departure $T$ . First, ferry 1: $a_{1}=v_{1}(8.20-T)$ and ferry 2: $a_{2}=v_{2}(8.20-T)$ ; sum $a=a_{1}+a_{2}=(v_{1}+v_{2})(8.20-T)$ . Second, difference in time: $9.11-8.20-0.15=0.36$ . Ferry 1: $a_{2}+a_{4}=v_{1}\cdot (0.36)$ and ferry 2: $a_{1}+a_{3}=v_{2}\cdot (0.36)$ ; sum $2a=(v_{1}+v_{2})(0.36)$ . Dividing: $\frac{1}{2}=\frac{8.20-T}{0.36}$ , so $T=8.02$ .

1997 PMWC (Problems)
Preceded by Problem T3	Followed by Problem T5
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

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1997 PMWC Problems/Problem T4

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