Question 192729
Jim drives from the Twin Cities to Monroe Louisiana, a distance of 1075 miles
 in record time. On the way back he averages 5 mph slower and the trip takes
 an additional 1.5 hrs. How fast did he drive to Monroe LA.?
 Round to the nearest tenth of a mph.
:
let s = his original speed to La
then
(s-5) = his return speed
;
Write a time equation: Time = {{{dist/speed}}}
;
Time to + 1.5 hr = time to return
{{{1075/s}}} + 1.5 = {{{1075/((s-5))}}}
Multiply equation by s(s-5), results
1075(s-5) + 1.5s(s-5) = 1075s
:
1075s - 5375 + 1.5s^2 - 7.5s - 1075s = 0
:
1.5s^2 - 7.5s - 5375 = 0
:
A quadratic equation, use the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation: x = s; a = 1.5, b = -7.5, c = -5375
{{{s = (-(-7.5) +- sqrt(-7.5^2 - 4 * 1.5 * -5375 ))/(2*1.5) }}}
:
{{{s = (7.5 +- sqrt(56.25 - (-32250) ))/(3) }}}
:
{{{s = (7.5 +- sqrt(56.25 + 32250 ))/(3) }}}
:
{{{s = (7.5 +- sqrt(32306.25 ))/(3) }}}
Two solutions but we only want the positive one:
{{{s = (7.5 + 179.74)/(3) }}}
{{{s = 187.24/3 }}}
s = 62.4 mph to Monroe La
:
:
Check solution by finding the time for each trip; return speed: 62.4 - 5 = 57.4
1075/57.4 = 18.73 hrs
1075/62.4 = 17.23 hrs
-----------------------
difference: 1.5 hrs