Question 53778: Can someone help me on this problem?
A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi. apart?
Thanks,
Sher
Found 2 solutions by venugopalramana, jenrobrody:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Can someone help me on this problem?
A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi. apart?
Thanks,
Sher
RELATIVE SPEED OF 2 TRAINS GOING IN OPPOSITE DIRECTIONS = 44+48=92 MPH
INITIAL DISTANCE OF SEPERATION = DISTANCE TRAVELLED IN 1 HR BY THE FIRST BUS = 44 MILES
FINAL DISTANCE OF SEPERATION = 274 MILES
HENCE RELATIVE DISTANCE TRAVELLED AFTER SIMULTANEOUS START = 274-44=230
TIME NEEDED = 230/92 HRS = 2.5 HRS
HENCE AT 1+2.5+1=00 PM = 4.5 PM THE 2 BUSES WILL BE 274 MILES APART
Answer by jenrobrody(19)  (Show Source):
You can put this solution on YOUR website! The first bus travels at 44mph for x hours.
The second bus travels at 48mph for (x-1) hours, it leaves an hour later-or travels one hour less than the first bus.
The distance the first bus travels (from D=RT), is 44x
The distance the second bus travels is 48(x-1), 48mph times (x-1) hours
Since the buses are traveling in different directions, the sum of their distances should equal 274. or in Algebra:
44x + 48(x-1) = 274 , then distribute:
44x + 48x - 48 = 274 , combine like terms:
92x - 48 = 274 , add 48 to both sides:
92x = 322 , divide by 92:
x = 3.5
So, the first bus travels 44 mph for 3.5 hours (a total distance of 154 miles).
The second bus travels 48 mph for 2.5 hours (3.5-1) for a distance of 120 miles.
Thus, 3.5 hours after the first bus leaves, ie. at 4:30pm, the two buses will be 274 miles apart.
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