<PRE><B>Question 370418</B>
Two cars each travel 75 miles at a constant rate.
 one car travels 6 mph faster than the other and arrives 5 minutes before the other arrives.
 Find the rates of speed of the two cars.
:
Let s = speed of the slower car
then
(s+6) = speed of the faster car
:
Convert 5 min to hrs 5/60 = 1/12 hr
:
Write a time equation: Time = dist/speed
:
Slow car time - faster car time = 5 min (1/12 hr)
{{{75/s}}} - {{{75/((s+6))}}} = {{{1/12}}}
:
Multiply by 12s(s+6) to get rid of the denominators, results:
12(s+6)*75 - 12s(75) = s(s+6)
:
900(s+6) - 900s = s^2 + 6s
:
900s + 5400  - 900s = s^2 + 6s
:
5400 = s^2 + 6s
:
A quadratic equation
0 = s^2 + 6s - 5400
:
Use the quadratic formula to find s
{{{s = (-6 +- sqrt(6^2-4*1*-5400 ))/(2*1) }}}
do the math, you should get a positive solution s ~ 70.546 mph


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