SOLUTION: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.

Question 80688: Steve traveled 200 miles at a certain speed. Had he gone 10mph 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; t=d/r and r=d/t
let r=initial speed
travel time at initial speed =200/r
travel time at 10 mph faster=200/(r+10)
Now we are told that the travel time at 10 mph faster (200/(r+10)) is 1 hour less than the travel time at the initial speed (200/r). So our equation to solve is:
(200/r)-1=200/(r+10) multiply both sides by r(r+10) to get rid of fractions
200(r+10)-r(r+10)=200r get rid of parens
200r+2000-r^2-10r=200r subtract 200r from both sides
200r-200r+2000-r^2-10r=200r-200r collect like terms
2000-r^2-10r=0 multiply each term by -1
r^2+10r-2000=0 quadratic in standard form and it can be factored.
(Note: If the A coefficient of a quadratic is 1, then the B coefficient will be the sum of the factors of the C coefficient. In this case, the factors of -2000 that add up to +10 are +50 and -40)
(r+50)(r-40)=0
r=-50 mph-------------------------discount negative speed
and
r=40 mph-------------------------------initial speed
Check
200/40=5 hr
200/50=4 hr
5 hr - 4 hr =1 hr

Hope this helps---ptaylor

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