HELP As part of his training an athlete usually runs 80km at
HELP!!!
 As part of his training, an athlete usually runs 80km at a steady speed of v km/. One day he decideds to reduce his speed by 2.5km/h and his run takes him an extrax 2h 40 min. Derive an equation in terms of v and use to find athlete\'s initial speed.
Solution
Before change DATA:
 distance = 80km ; rate = v km/h ; time = d/r = 80/v hrs.
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 After change DATA:
 distance = 80km ; rate = (v-2.5)km/h ; time = d/r = 80/(v-2.5) hrs
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 EQUATION:
 After time - Before time = 2 2/3 hr.
 80/(v-2.5) - 80/v = 8/3
 Divide thru by 8 to get:
 10/(v-2.5) - 10/v = 1/3
 Multiply thru with 3v(v-2.5) to get:
 30v - 30(v-2.5) = v(v-2.5)
 2.5 = v^2-2.5v
 v^2-2.5v-2.5 = 0
 v = [2.5 +- sqrt(6.25 - 4*-2.5)]/2
 v = [2.5 +- sqrt(6.25 + 10)]/2
 Positive solution:
 v = [2.5 + 4.03]/2
 v = 3.26556....km/h

