SOLUTION: If two groups of cyclists, 4 minutes apart, are both going in the same direction and are both riding at 20km/h, at what speed will one cyclist from the second group have to ride to

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Question 617072: If two groups of cyclists, 4 minutes apart, are both going in the same direction and are both riding at 20km/h, at what speed will one cyclist from the second group have to ride to catch the first group within 6km?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
4 minutes apart with both groups riding at 20 km/hr
means they are +20%2A%284%2F60%29+=+4%2F3+ km apart
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Let +s+ = the additional speed in km/hr needed to make up the
4/3 km in +t%2F60+ hrs
Relative to the moving groups:
(1) +s%2A%28t%2F60%29+=+4%2F3+ km
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Relative to the ground:
(2) +%28+20+%2B+s+%29%2A%28t%2F60%29+=+6+ km
+t%2F3+%2B+s%2A%28t%2F60%29+=+6+
Substitute (1) into (2)
+t%2F3+%2B+4%2F3+=+6+
multiply both sides by +3+
+t+%2B+4+=+18+
+t+=+18+-+4+
+t+=+14+ min
and
(1) +s%2A%28t%2F60%29+=+4%2F3+
(1) +s%2A%2814%2F60%29+=+4%2F3+
(1) +14s+=+80+
(1) +s+=+80%2F14+
(1) +s+=+5.714+
+20+%2B+5.714+=+25.714+
A cyclist from the 2nd group will have
to ride at 25.714 km/hr
check:
(2) +%28+20+%2B+s+%29%2A%28t%2F60%29+=+6+ km
(2) +25.714%2A%2814%2F60%29+=+6+
(2) +25.714%2A14+=+360+
(2) +359.996+=+360+
close enough