Question 1160245: Jacques left Sachin's house yesterday walking homewards at 6 kilometers per hour. After 19 minutes, he realized that he had left his mobile phone behind.
Immediately, he turned round to walk back, stepping up his pace to 9 kilometers per hour, unaware that Sachin had already been running at 15 kilometers per hour for 2 minutes, trying to catch him. They continued at 9 kilometers per hour and 15 kilometers per hour respectively until they encountered each other. How far from Sachin's house did they encounter each other? I don't know how to do this, I've tried converting kilometers to meters then to minutes. It's somewhere between 1025 and 1500 meters I think.
Answer by greenestamps(13209)   (Show Source):
You can put this solution on YOUR website!
The distance Jacques traveled before he turned around was
19 minutes at 6km/hr
= 19 minutes at 6000m/hr
= 19 minutes at 100m/min
= 1900m
At the time he turned around the distance Sachin had already traveled was
2 minutes at 15km/hr
= 2 minutes at 15000m/hr
= 2 minutes at 250m/min
= 500m
So the distance between them when Jacques turned around was 1900-500 = 1400m.
After that, they were traveling toward each other at a rate of
9km/hr+15km/hr = 24km/hr
= 24000m/hr
= 400m/min
The additional time it would take them to meet is
1400m/(400m/min) = 3.5 minutes
The distance from Sachin's house that they met is the distance Sachin runs in 2+3.5=5.5 minutes at 250m/min = 1375m.
ANSWER: 1375 meters
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