A car leaves the Kansas City at noon traveling west at 45 mi

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?

Solution

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.