Question 250844: A car leaves the Kansas City at noon traveling west at 45 miles per hour. A second car leaves from the same location 3 hours later traveling west at 60 miles per hour. At what time will the second car catch up with the first?
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! We need to set up an RTD chart to solve this:
west . . . . . RATE . . . . . time . . . . distance
CAR 1 . . . . 45 mph . . . . . t . . . . . 45t
CAR 2 . . . . 60 mph . . . . . t-3 . . . . 60t - 180
we are given car 1 and car 2 rates as 45 and 60 respectively. Car 1 leaves first and car 2 3 hours later. This means that car 2 is on the road for 3 fewer hours, t-3. rate x time = distance. This is where I got 45t and 60t - 180. For the second car to catch up to the first means the distances must be equal.
So,
45t = 60t - 180.
Solving for t, we get
T = 12 hours.
Since car 1 left at noon (12 o'clock), the second car would catch up at midnight.
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