a bus leaves a station at 1 p.m., traveling west at an average rate of 44mph. one hour later a second bus leaves the same station, traveling east at a rate of 48mph. At what time will the two buses be 274 miles apart? Please I can not figure it out.
2006-11-24 21:26:53 · 7 answers · asked by Tasha J 1 in Science & Mathematics ➔ Mathematics
At 1 p.m the 1st bus starts moving west with a velocity of 44 mph that means at 2 pm the bus will be 44 miles away from the station because D = V x t = 44 mph x 1 h = 44 miles .
At 2 pm the 2nd bus starts moving east i.e in the opposite direction with a velocity of 48 mph
B1 ____________B2 ( B1 B2 = 44 miles )
After 2 pm the distance between the 2 buses at any time is equal
to :
D = 44 +( V1 x t ) + (V2 x t ) = 44 + (44 x t ) + (48 x t ) = 44 + 92 t
where D >> miles , t >> hrs
B1 _________S___________ B2
Where B1 S = 44 miles + 44 t
B2 S = 48 t
Put D = 274 miles to get t
274 = 44 + 92 t >> t = 2.5 h
Therefore the two buses will be a distance of 274 miles apart at :
2 pm + 2.5 h = 4 : 30 pm
2006-11-25 02:43:28 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Sounds like you need to solve a system of equations.
First off, the first bus: It's distance can be represented by the following:
d1 = 44t + 44 (standardizing it to the time when the second bus leaves)
The second bus can be represented as:
d2 = -48t
Then you need to solve the equations so that the difference between the two positions is 274 miles.
d1-d2 = 274 = (44t+44) - -48t = 92t + 44
274 - 44 = 92t, t = 2.5 hours
Add this to 1400 hours, the time bus two left, and you have 1630 hours. Or, 4:30 p.m
That is, unless I messed something up bad. Heh.
2006-11-25 05:32:37 · answer #2 · answered by DJL2 3 · 0⤊ 0⤋
Its not the pretty equation you were after, but here is the answer:
After 1 hour distance (d) = 44 miles
After 2 hours d = 44x2 + 48 = 136 miles
After 3 hours d = 44x3 + 48 x 2 = 228 miles
After 3.5 hours d = 228 + 22 + 24
d = 274 miles.
1 pm + 3.5 hours = 4.30 pm
2006-11-25 05:35:43 · answer #3 · answered by 13caesars 4 · 0⤊ 0⤋
the distance travelled by bus1 after an hour=44 km
the distanceto be put in by bothe the buses
=274km-44km=230 km
relative speed=44+48=92 kmph
time=distance/speed=230/92
=2 1/2 hours
so they will be 274 km apart at 4.30 p.m.
2006-11-25 05:30:20 · answer #4 · answered by raj 7 · 0⤊ 0⤋
Let time taken be x
First bus covers
44x
Second bus covers 48(x-1)
So, 44+48(x-1) = 274
or 48x = 270
or x = 45/8 = 5.625 hours
That is around 6:30 pm
2006-11-25 05:32:01 · answer #5 · answered by ag_iitkgp 7 · 0⤊ 0⤋
5.pm
2006-11-25 05:44:54 · answer #6 · answered by lashawn t 1 · 0⤊ 0⤋
Is this for your test or somethin. 2hr 33min
2006-11-25 05:29:29 · answer #7 · answered by brilliantrobert 4 · 0⤊ 0⤋