The cyclist set off at a constant speed from city A to city B, the distance between which is 120 km.

| June 11, 2021 | education

The cyclist set off at a constant speed from city A to city B, the distance between which is 120 km. The next day, he set off back at a speed of 5 km / h more than before. On the way, he made a stop for 4 hours. As a result, he spent the same amount of time on the return trip as on the path from A to B. Find the speed of the cyclist on the path from A to B.

Suppose a cyclist was traveling from city A to city B at a speed of x km / h, then the travel time was 120 / x hours.
He drove back 120 / (x + 5) hours. Let’s compose and solve the equation taking into account the parking time.

120 / x = 120 / (x + 5) + 4.

120 (x + 5) = 120x + 4x (x + 5).

120x + 600 = 120x + 4x ^ 2 + 20x.

Here are similar ones.

4x ^ 2 + 20x – 600 = 0.

Reduce the quadratic equation by 4 and find its root.

x ^ 2 + 5x – 150 = 0.

D = 25 + 4 * 150 = 625 = 25 ^ 2.

x = (- 5 + 25) / 2 = 20/2 = 10 km / h.

Answer: on the way from A to B, the cyclist rode at a speed of 10 kilometers per hour.



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