SOLUTION: A bus traveling at a constant rate of 30 miles per hour left the city at 11:45 a.m. A car following the bus at 45 miles per hour left the city at noon. At what time did the car ca

Algebra ->  Expressions-with-variables -> SOLUTION: A bus traveling at a constant rate of 30 miles per hour left the city at 11:45 a.m. A car following the bus at 45 miles per hour left the city at noon. At what time did the car ca      Log On


   

Question 352513: A bus traveling at a constant rate of 30 miles per hour left the city at 11:45 a.m. A car following the bus at 45 miles per hour left the city at noon. At what time did the car catch up with the bus?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let t = the amount of time the car would catch up with the bus. The car took off 15 minutes, or 1/4 of an hour after the bus. Therefore the amount of time before the car catches up with the bus is t-1%2F4. From the formula
D+=+RT (or distance = rate x time),
45%2A%28t+-+1%2F4%29=30%2At,
45%2At-45%2F4+=+30%2At,
15%2At+=+45%2F4,
t+=+3%2F4.
So the car would catch up with the bus after 3/4 hr, or 45 minutes.