SOLUTION: Art leaves his house at 10 o'clock traveling at 50 mph. Jennifer leaves her house at 10:30 traveling at 60 mph. At what time will Jennifer catch up to Art?
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Question 823617: Art leaves his house at 10 o'clock traveling at 50 mph. Jennifer leaves her house at 10:30 traveling at 60 mph. At what time will Jennifer catch up to Art? Answer by algebrahouse.com(1659) (Show Source):
Art's distance
D = rt {distance = rate x time}
rate = 50
time = t
d = 50t
Jennifer's distance
D = rt {distance = rate x time}
rate = 60
time = t - 0.5 {left 1/2 hour later than Art}
d = 60(t - 0.5) = 60t - 30 {used distributive property}
50t = 60t - 30 {when Jennifer catches up with Art, the distances will be equal}
-10t = -30 {subtracted 60t from each side}
t = 3 {divided each side by -10}
