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 cat

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Question 834823: 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 josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Quarter of an hour difference. Car drives for time t but bus drives for time t%2B1%2F4.

_________________speed_________time_________distance
Bus_____________30_____________t+1/4________d
Car_____________45_____________t____________d

Know again, t is not a point on a number time line; it is a quantity of time, in hours.

The situation has these equations:
30%28t%2B1%2F4%29=d and 45t=d.
The unknown variables are t, the time for car to catchup distance, and d, that distance. Since two expressions using the variable to be solved are both equal to d, we can equate these expressions in an equation:

highlight_green%2830%28t%2B1%2F4%29=45t%29
Solve for t.
Use t to find the time of day that the two vehicles reach the same distance.