SOLUTION: a Tour bus leaves Sacramento every Friday evening at 5 pm for a 170 mile trip. this week however the bus leaves at 5:30 pm. to arrive on time , the driver drives 9 miles per hour

Question 517775: a Tour bus leaves Sacramento every Friday evening at 5 pm for a 170 mile trip. this week however the bus leaves at 5:30 pm. to arrive on time , the driver drives 9 miles per hour faster than usual what is the buses usual speed? please show work
Found 3 solutions by richwmiller, josmiceli, solver91311:
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
r*t=d
r*t=170
(r+9)*(t-.5)=170
r=51
r+9=60
60*(t-.5)=170
60t-30=170
60t=200
divide by 20
3t=10
t=10/3=3 hrs 20 minutes
or
51t=170
t=170/51=3.333

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = the usual driving time
in hours to to arrive on schedule
Let = the bus's usual speed in mi/hr
--------------
given:
For the usual trip:
(1)
For the late trip:
(2)
-----------------------
(2)
Substitute (1) into (2)
(2)
(2)
also
(1)
(2)
Multiply both sides by
(2)
(2)
Use quadratic equation

and, from (1)
(1)
(1)
(1)
The usual speed is 51 mi/hr
check answer:
(2)
(2)
(2)
(2)
(2)
OK
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Let represent the usual speed. Let represent the usual amount of time. Then represents the increased speed and represents the reduced time.

Time is equal to distance divided by rate so for the normal trip:

And for the 'catch-up' trip:

Which is to say:

A little Algebra Music, Maestro:

Now that we have two expression that are each equal to , set them equal:

All that is left is to cross multiply, collect terms, and solve for

John

My calculator said it, I believe it, that settles it

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