The train had to travel 840 km. After passing a third of the way at the speed provided by the schedule, he was delayed for 1 hour at the semaphore, and in order to arrive at the destination on the schedule, he had to increase the speed by 10 km / h. At what speed should the train travel?
Stot = 840
S1 = 1 / 3Stotal = 280 – distance of the 1st track section
S2 = 2 / 3Stotal = 560 – distance of the 2nd track section
x – planned speed
x + 10 – increased speed
t – total travel time
t = Stot / x = 840 / x
t1 + t2 + 1 = t
t1 = S1 / x = 280 / x – time of passage of the 1st section of the path
t2 = S2 / x + 10 = 560 / x + 10 – time of passage of the 2nd section of the path
280 / x + 560 / (x + 10) + 1 = 840 / x
560 / (x + 10) + 1 = 560 / x
560 / x = (560 + x + 10) / (x + 10)
560x + 5600 = x ^ 2 + 570x
x ^ 2 + 10x-5600 = 0 – quadratic equation
We solve it:
D = b ^ 2-4ac = 100-4 * 1 * 5600 = 22500
x = (- b + √D) / 2a = (- 10 + 150) / 2 = 70 km / h – problem solution
p.s. the second solution of the quadratic equation is not taken into account because speed based on the task cannot be negative.
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