Question 419295: Kansas City is 500 miles from Nashville. Sandy left Kansas City, traveling toward Nashville at an average speed at 50 miles per hour. One hour later, Shelly left Nashville travleing toward Kansas City at an average speed of 40 miles per hour. How long after Sandy left Kansas City will they meet?
Answer by duckness73(47)   (Show Source):
You can put this solution on YOUR website! Remember that Distance = Rate * Time (abbreviated: D = R * T)
Let t = time after Sandy left Kansas City that she will meet Shelly. Note that since Shelly left one hour later, she travelled one hour less, so Shelly will meet up with Sandy at t - 1.
Let d = distance that Sandy drove until she met up with Shelly. Note that since the distance between Kansas City and Nashville is 500 miles, that Shelly will have driven 500 - d when she meets up with Sandy.
Using the distance = rate * time formula, at the time that the two meet, we have:
Sandy travelled d = 50t
Shelly travelled 500 - d = 40(t - 1)
Putting the two equations together by substituting 50t from the first equation into the second equation, we have:
500 - 50t = 40(t - 1)
500 - 50t = 40t - 40
540 - 50t = 40t (adding 40 to both sides)
540 = 90t (adding 50t to both sides)
6 = t (dividing both sides by 90)
So, Sandy travelled 6 hours and covered 300 miles when she met up with Shelly who travelled 5 hours and covered 200 miles.
Answer: 6 hours
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