From point A to point B, located 24 km from A. The cyclist arrived at point B 4 hours earlier than walking

From point A to point B, located 24 km from A. The cyclist arrived at point B 4 hours earlier than walking. It is true that if the cyclist was traveling at a lower speed by 4 km / h, then the path from A to But it would take half the time that the walker would. Find the speed of the walker.

Let the speed of the cyclist be V1 and the speed of the pedestrian V2 km / h.

Then the cyclist spent time on the road from point A to point B:

t1 = S / V1 / = 24 / V1, where S is the distance from A to B.

The pedestrian spent time:

t2 = S / V2 = 24 / V2.

By condition t2 – t1 = 4 hours.

(24 / V2) – (24 / V1) = 4.

24 / V2 = 4 + (24 / V1). (one).

If the cyclist’s speed is (V1 – 4), then t1 = t2 / 2.

24 / (V1 – 4) = (24 / V2) / 2.

24 / V2 = 2 * 24 / (V1 – 4) (2).

Let’s solve the system of two equations 1 and 2.

4 + (24 / V1) = 48 / (V1 – 4).

(4 * V1 + 24) * (V1 – 4) = 48 * V1.

4 * V12 + 24 * V1 – 16 * V1 – 96 – 48 * V1 = 0.

4 * V12 – 40 * V1 – 96 = 0.

V12 – 10 * V1 – 24 = 0.

Let’s solve the quadratic equation.

D = b2 – 4 * a * c = (-10) 2 – 4 * 1 * (-24) = 100 + 96 = 196.

X1 = (10 – √196) / (2 * 1) = (10 – 14) / 2 = -4 / 2 = -2. (Doesn’t fit because <0).

X2 = (10 + √196) / (2 * 1) = (10 + 14) / 2 = 24/2 = 12.

V1 = 12 km / h.

24 / V2 = 4 + (24 / V1). (one).

V2 = 24 / (4 + 24/12) = 24/6 = 4 km / h.

Answer: Pedestrian speed is 4 km / h.