SOLUTION: A tour bus normally leaves for its destination at 5:00 p.m. for a 330 mile trip. This week however, the bus leaves at 5:30 p.m. To arrive on time, the driver drives 5 miles per hou

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A tour bus normally leaves for its destination at 5:00 p.m. for a 330 mile trip. This week however, the bus leaves at 5:30 p.m. To arrive on time, the driver drives 5 miles per hou      Log On


   

Question 959641: A tour bus normally leaves for its destination at 5:00 p.m. for a 330 mile trip. This week however, the bus leaves at 5:30 p.m. To arrive on time, the driver drives 5 miles per hour faster than usual. What is the bus` usual speed?
I did my best, it was not good enough
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the usual speed in mi/hr that will result
in the bus arriving on time
Let +t+ = the time in hours that the trip takes
when the bus has speed +s+
----------------------------------------------
Equation for a normal trip:
(1) +330+=+s%2At+
Equation for trip leaving at 5:30 PM
(2) +330+=+%28+s+%2B+5+%29%2A%28+t+-+.5+%29+
---------------------------------
(2) +330+=+s%2At+%2B+5t+-+.5s+-+2.5+
(2) +332.5+=+s%2At+%2B+5t+-+.5s+
and
(1) +s+=+330%2Ft+
---------------------
By substitution:
(2) +332.5++=+330+%2B+5t+-+.5%2A%28+330%2Ft+%29+
(2) +2.5+=+5t+-+165%2Ft+
(2) +5t%5E2+-+2.5t+-+165+=+0+
Divide both sides by +2.5+
(2) +2t%5E2+-+t+-+66+=+0+
Use quadratic formula
+t+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+2+
+b+=+-1+
+c+=+-66+
+t+=+%28+-%28-1%29+%2B-+sqrt%28+%28-1%29%5E2+-+4%2A2%2A%28-66%29+%29%29+%2F+%282%2A2%29+
+t+=+%28+1+%2B-+sqrt%28+1+%2B+528+%29%29+%2F+4+
+t+=+%28+1+%2B-+sqrt%28+529+%29%29+%2F+4+
+t+=+%28+1+%2B+23+%29+%2F+4+
+t+=+24%2F4+
+t+=+6+
and
(1) +s+=+330%2Ft+
(1) +s+=+330%2F6+
(1) +s+=+55+
The usual speed is 55 mi/hr
check:
(2) +330+=+%28+s+%2B+5+%29%2A%28+t+-+.5+%29+
(2) +330+=+%28+55+%2B+5+%29%2A%28+6+-+.5+%29+
(2) +330+=+60%2A5.5+
(2) +330+=+330+
OK