Question 132829: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
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
Disregard the negative value for r. In this case speed is positive
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
