Distance: A bus traveling at an average rate of 30 mph left

A bus traveling at an average 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?

let t be the time in hours since bus left the station
distance D the bus travelled is 30t

the time the car traveled is [ t-1/4] because 15 minutes is 1/4 an hour
distance car traveled D=45[t-1/4]

the distances are the same when D=D or
30t=45[t-1/4] clear bracket
30t=45t-45/4 subtract 30t from each side of = sign
0=15t -45/4 add 45/4 to each side
45/4 = 15t divide each side by 15
3/4 =t or they meet 45 minutes after the bus left or at 12:30


another way
the car travels 15 mph faster than the bus
the bus traveled 15 minutes[or 1/4 hour] longer than the car or 30/4 miles more
it takes the car [30/4] / 15 hours to catch the bus or 1/2 hour after the car left at 12 o clock
the car reaches the bus at 12:30

Arthur