Posted by Dani on Sunday, November 30, 2008 at 12:02pm.
let his first steady speed by x mph
so the time for the first leg of the trip was 120/x hours
on the second leg of the trip his speed was x+10 mph
so the time taken for that was 150/(x+10)
but this took 6 min or 1/10 hour longer
so 150/(x+10) - 120/x = 1/10
multiply each term by x(x+10) to get a quadratic equation.
one of the roots should be extraneous, let me know what you got
Let the two speeds be V1 and V2. The times required to travel the two legs of the journey are
120/V1 and 150/V2 = 150/(V1+10)
The second leg took 0.1 hours (6 minutes) more, so
150/(V1+10) = (120/V1) + 0.1
That equation can be solved for V1.
150/(V1 + 10) = (120 + 0.1V1)/V1
150 V1 = 120 V1 + 1200 + V1 + 0.1V1^2
0.1V1^2 -29 V1 + 1200 = 0
V1 = [29 +/-19]/0.2 = 50 or 240 mph
The higher of the two roots is unrealistic. Go with 50 mph
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