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Difference between revisions of "1997 PMWC Problems/Problem T4"

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==Problem==
 
==Problem==
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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?  
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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 <math>8:20</math>. The two ferries then sailed to their destinations, stopped for <math>15</math> minutes and returned. The two ferries met again at <math>9:11</math>. Suppose the two ferries traveled at a uniform speed throughout the whole journey, what time did the two ferries start their journey?
  
 
==Solution==
 
==Solution==
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Time <math>t</math>, velocity <math>v</math>, distance <math>a</math>, formula <math>v=\frac{a}{t}</math> or <math>a=v \cdot t</math>.
 
Time <math>t</math>, velocity <math>v</math>, distance <math>a</math>, formula <math>v=\frac{a}{t}</math> or <math>a=v \cdot t</math>.
 
Time of departure <math>T</math>.
 
Time of departure <math>T</math>.

Latest revision as of 13:40, 20 April 2014

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 $T=8.02$.

See Also

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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