SOLUTION: 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 along the same route, traveling at 54 mi/h. At what time will t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: 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 along the same route, traveling at 54 mi/h. At what time will t      Log On


   

Question 43220: 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 along the same route, traveling at 54 mi/h. At what time will the two buses be 274 mi apart?
Found 2 solutions by aaaaaaaa, AnlytcPhil:
Answer by aaaaaaaa(138) About Me  (Show Source):
You can put this solution on YOUR website!
The position of the first bus at x hours (counting from the time the second bus departed) is:
d%5B1%5D+=+44x+%2B+44
The position of the second one is:
d%5B2%5D+=+54x
We want to solve for how many hours (x), the following equation will be true:
d%5B2%5D-d%5B1%5D+=+274
Since we used d%5B1%5D and d%5B2%5D only for clarity, let's substitute for the true formulas:
54x+-+%2844x+%2B+44%29+=+274
10x+-+44+=+274
And we have a linear equation! Solving it we have:
Solved by pluggable solver: SOLVE a linear equation
Solve 10%2Ax%2B-44=274. Move -44 to the right: 10%2Ax=274--44. Divide by 10: x=%28274--44%29%2F10=318%2F10+=+31.8


Verifying...
54%2A%2831.8%29+=+1717.2
44%2A%2831.8%29+%2B+44+=+1443.2
1717+-+1443+=+274
Therefore, that distance will happen at 31.8 hours after the second bus left (2 p.m or 14:00). Remember that 0.8 hours is 0.1%2A60%2A8 minutes, or 48. To get to 24:00 we need 10 hours, so it will happen at 31 - 10 = 21 hours and 48 minutes of the next day. A long time!

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
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 along the 
same route, traveling at 54 mi/h.  At what time will the two 
buses be 274 mi apart?

Wow!!!!!!  That'll sure take a long time with the second bus 
traveling only 10 mi/hr faster than the first bus!!!!!!! At
only 10 mi/hr faster, the second bus won't even catch up to
the first to pass it until 4.4 hours later at 6:24 PM.  Then
to get 274 miles ahead will take more than a whole day.  Are
you sure you have this one copied right?  It's a ridiculous
problem, but it can be imagined. So I'll do it anyway.  Let's
assume they started on Monday.  It will be at least Tuesday 
evening before the 2nd bus is 274 miles ahead of the first bus,
going only 10 mph faster.  

Make this chart

             D      R    T   
1st bus|    
2nd bus| 

Let t be the time the first bus traveled.
Then since the 2nd bus started an hour later
its traveling time is t-1.  So fill in the 
times

             D      R    T   
1st bus|                 t
2nd bus|                t-1

Fill in the given rates (speeds):

             D      R    T   
1st bus|           44    t
2nd bus|           54   t-1

Use D = RT to fill in the distances:

             D      R    T   
1st bus|    44t    44    t
2nd bus|  54(t-1)  54   t-1

Now the distance the 2nd bus traveled
is 274 miles more than the 1st bus

So we form the equation by

2nd bus's distance = 1st bus's distance + 274

   54(t-1) = 44t + 274 

Answer = 32.8 hours

So if the first bus started at 1 p.m on Monday
the second bus will be 274 miles ahead of the
1st bus at 9:48 p.m Tuesday night.

I think you copied the problem wrong.

Edwin