# From point A to point B, the distance between which is 20 km, a pedestrian left. Simultaneously with it, a cyclist left

**From point A to point B, the distance between which is 20 km, a pedestrian left. Simultaneously with it, a cyclist left point B to point A and met a pedestrian 50 minutes after leaving B. How long would it take for a pedestrian to go all the way from A to B if it is known that the cyclist would have done the same the way is 4 hours faster than a pedestrian?**

Given:

S = S1 + S2 = 20 km

S1 = vp * Tv

S2 = vw * Tw

TV = 50 min = 5/6 h – time before meeting

(5/6) * vp + (5/6) * vv = 20

5 * (vp + vv) = 120

vp + vv = 24

vv = 24-vp

also from the conditions of the assignment it is known:

tp = tv + 4

S = vp * tp

S = vw * tv = vw * (tp-4)

vp * tp = vw * (tp-4)

earlier it was found out vv = 24-vp, therefore

vp * tp = (24-vp) * (tp-4)

vp = S / tp = 20 / tp, then we solve the following expression:

20 / tp * tp = (24-20 / tp) * (tp-4)

20 = 24 * tp-20-96 + 80 / tp

0 = 24 * tp-136 + 80 / tp

80 / tp = 136-24 * tp

80 = 136 * tp-24 * tp ^ 2

10 = 17 * tp-3 * tp ^ 2

3 * tp ^ 2-17 * tp + 10 = 0

we solve the quadratic equation:

D = (- 17) ^ 2-4 * 3 * 10 = 169

tp1 = – (- 17) + √169 / 2 * 3 = (17 + 13) / 6 = 5h

tp2 = – (- 17) -√169 / 2 * 3 = (17-13) / 6 = 2 / 3h

Of the two solutions, the 5 o’clock answer seems to be more appropriate, since from the conditions of the problem, a cyclist travels a given distance 4 hours faster than a pedestrian, and the time cannot be negative.

Answer: 5 hours