Asked by allie
                diana drove 320 miles, then increased her speed by 10 miles per hour for 500 miles. if the second part of the trip took 2 hours longer than the first, find her average speed.
            
            
        Answers
                    Answered by
            Reiny
            
    let her initial speed be x mph
time for first trip = 320/x
new speed = x+10 mph
time for second part of trip = 500/(x+10)
so..
500/(x+10) - 320/x = 2
times x(x+10)
500x - 320(x+10) = 2x(x+10)
500x - 320x - 3200 = 2x^2 + 20x
2x^2 - 160x + 3200 = 0
x^2 - 80x + 1600 = 0
(x-40^2 = 0
x = 40
time for first leg = 320/40 =8 hrs
time for 2nd leg = 500/50 = 10 hrs
avg speed = total distance/total time
= 820/18 = appr 45.56 mph
    
time for first trip = 320/x
new speed = x+10 mph
time for second part of trip = 500/(x+10)
so..
500/(x+10) - 320/x = 2
times x(x+10)
500x - 320(x+10) = 2x(x+10)
500x - 320x - 3200 = 2x^2 + 20x
2x^2 - 160x + 3200 = 0
x^2 - 80x + 1600 = 0
(x-40^2 = 0
x = 40
time for first leg = 320/40 =8 hrs
time for 2nd leg = 500/50 = 10 hrs
avg speed = total distance/total time
= 820/18 = appr 45.56 mph
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