SOLUTION: Leah leaves work at 17:54, halfway home she meets her daughter who left home at 17:46. Knowing that Leah walks 1.5x faster than her daughter and that they are travelling in a strai

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Question 941053: Leah leaves work at 17:54, halfway home she meets her daughter who left home at 17:46. Knowing that Leah walks 1.5x faster than her daughter and that they are travelling in a straight line, at what time do they meet?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = the distance in feet from work to home
Find Leah's head start:
17:54 minus 17:46 = 8 min
Let +s+ = the daughter's walking speed in ft/min
+1.5s+ = Leah's walking speed
+d%5B1%5D+=+1.5s%2A8+
+d%5B1%5D+=+12s+ ft
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Start a stop watch when daughter leaves
add their speeds to find when they meet
+d+-+12s+=+%28+s+%2B+1.5s+%29%2At+
+t+=+%28+d+-+12s+%29+%2F+%282.5s%29+
+t+=+d%2F%282.5s%29+-+4.8+ minutes
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They meet at 17:46 + d/(2.5s) - 4.8