The donkey, having passed the bridge 3/8 of its length, looked back and saw a car moving towards the bridge.

The donkey, having passed the bridge 3/8 of its length, looked back and saw a car moving towards the bridge. If the donkey turns back, it will meet a car at the beginning of the bridge, and if it runs forward, it will meet a car at the end of the bridge. With what speed did the frightened donkey run if the speed of the car was V?

Let the entire length of the bridge be S, and the distance from the car to the bridge, at the moment when the donkey saw it, is S1. Also, the donkey’s speed is v1. If the donkey runs back, then during the time that it reaches the beginning of the bridge, the car will also approach the bridge, that is:

3 * S / (8 * v1) = S1 / v

If the donkey runs forward, then during the time the donkey reaches the end of the bridge, the car will also pass the bridge, that is:

5 * S / (8 * v1) = S1 / v + S / v.

Substituting the first equation into the second, we get:

5 * S / (8 * v1) = 3 * S / (8 * v1) + S / v;

5 * S / (8 * v1) – 3 * S / (8 * v1) = S / v;

2 * S / (8 * v1) = S / v;

2 / (8 * v1) = 1 / v;

1 / (4 * v1) = 1 / v;

v = 4 * v1;

v1 = 1/4 * v = 0.25 * v.

Answer: the speed of the donkey is 1/4 * v (0.25 of the speed of the car).