SOLUTION:
A car leaves at 7a.m. on a long trip, traveling at 44kph. A second car leaves at 11 a.m. following the first car at 108kph. At what time does the second car catch the first?
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A car leaves at 7a.m. on a long trip, traveling at 44kph. A second car leaves at 11 a.m. following the first car at 108kph. At what time does the second car catch the first?
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Question 712082:
A car leaves at 7a.m. on a long trip, traveling at 44kph. A second car leaves at 11 a.m. following the first car at 108kph. At what time does the second car catch the first? Answer by checkley79(3341) (Show Source):
You can put this solution on YOUR website! D=RT
D=44T
D=108(T-4)
44T=108(T-4)
44T=108T-432
44T-108T=-432
-64T=-432
T=-432/-64
T=6.75 HOURS.
OR 7+6.75=13.75=1 HOUR 45 MIN. THEY WILL MEET.
PROOF:
44*6.75=108(6.75-4)
297=108*2.75
297=297
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