Question 330782: At 8:00am Dave leaves his home on his moped, traveling east at 18 mph. At the same time and 80 mile away, Cathy leaves her home on her bicycle, traveling west toward Dave at 14 mph. At what time will they meet?
Answer by texttutoring(324)   (Show Source):
You can put this solution on YOUR website! You know that D=VT, where D=distance, V=velocity (speed) and T = time.
Rearranging for time, we have D/V = T
We know that their times are the same, so Cathy's time = Dave's time
Tc = Td
Dc/Vc = Dd/Vd
Where Dc=Cathy's distance when they meet, Vc = Cathy's velocity, and so on...
We know that Dc+Dd = 80, (or Dc =80-Dd) because they start 80 miles apart and are heading towards each other.
Put in what you know:
Dc/14 = Dd/18
80-Dd/14 = Dd/18
Solve for Dd by cross multiplying:
18(80-Dd) = 14Dd
32Dd = 1440
Dd = 45
So Dave travels 45 miles. Now find his time:
T = Dd/Vd
T = 45/18
T = 2.5
So they will meet after 2.5 hours. You can solve for T in the equation for Cathy, and it will be the same.
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