SOLUTION: Please help me understand how to do this problem! Thanks sooo much in advance Here is the question: A bus traveling 40mi/h left Freetown at noon. A car following the bus at

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Please help me understand how to do this problem! Thanks sooo much in advance Here is the question: A bus traveling 40mi/h left Freetown at noon. A car following the bus at       Log On


   

Question 140739This question is from textbook Prentice Hall Mathmatics Pre-Algebra
: Please help me understand how to do this problem! Thanks sooo much in advance
Here is the question:

A bus traveling 40mi/h left Freetown at noon. A car following the bus at 60mi/h left Freetwon at 1:30 P.M. What time did the car catch up with the bus? How many miles were the car and the bus from Freetown whe the car caught up with the bus?
Thank you SO much in advance:) This question is from textbook Prentice Hall Mathmatics Pre-Algebra

Answer by solver91311(24713) About Me  (Show Source):
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The intuitive way to solve this one is to say that the bus had already traveled for an hour and a half at 40 miles per hour by the time the car left, so the bus was 1.5 times 40 miles or 60 miles from Freetown when the car left. The car has to use up 40 mph of its 60 mpm speed just to keep up with the bus, so it is only catching up with the bus at the difference in their speeds, or 20 mph. So how long does it take to go 60 miles at 20 miles per hour? Ans: 60/20 = 3 hours. 3 hours after the car left Freetown, it caught the bus, so the car traveled 3 hours times 60 mph = 180 miles.

The algebraic way to solve it is to remember that distance equals rate times time. So: d=rt

We don't know the total time driven by the car yet, so let's leave it expressed as t. We do know that the bus will have traveled for 1.5 hours longer than the car, so we can call the time for the bus t%2B1.5

We don't know the total distance traveled either, but we do know that the distance is the same for both the bus and the car -- once the car catches the bus, they will be in the same place.

And we know that r for the car is 60 mph, and r for the bus is 40 mph. So the distance traveled by the car is d=60t, and the distance traveled by the bus is d=40%28t%2B1.5%29. But remember, the distances are equal so:

60t=40%28t%2B1.5%29

60t=40t%2B60

60t-40t=60

20t=60

t=3

So the time the car was travelling was 3 hours.

Since we know for the car that d=60t, we can now say d=60%283%29=180

Check the answer:
The bus started out 1.5 hours earlier, so it must have been moving for 3 + 1.5 hours or 4.5 hours. So the distance traveled by the bus must be d+=+40%2A4.5+=+180. The distances are the same, so the car must have caught the bus.