Question 132829

Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r

Let r=the certain speed
And r+20=the 20 mph faster speed

Time required to go 600 mi at the certain speed=600/r
Time to go 600 mi at the faster speed=600/(r+20)

Now we are told that the time at this faster speed is one hour less than the time at the certain speed, so:

(600/r)-1=600/(r+20)  multiply each term by r(r+20)
600(r+20)-r(r+20)=600r  get rid of parens

600r+12000-r^2-20r=600r  subtract 600r from each side

600r-600r+12000-r^2-20r=600r-600r  collect like terms

-r^2-20r+12000=0  multiply each term by -1
r^2+20r-12000=0---------------------quadratic in standard form
Solve using the quadratic formula
{{{r= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{r = (-20 +- sqrt( 400+48000 ))/(2) }}} 
{{{r = (-20 +- sqrt( 52000 ))/(2) }}} 
{{{r = (-20 +-228.035)/(2) }}} 

Disregard the negative value for r.  In this case speed is positive

{{{r = (-20 +228.035)/(2) }}} 
{{{r = (-20 +228.035)/(2) }}} 
r=~~~~~~~104 mph


CK
Time at 104 mph=600/104=5.769 ~~~~~~5.8 hrs
Time at 104+20=600/124=4.8387 ~~~~~~~4.8 hrs


Hope this helps---ptaylor