When the passenger had to travel the distance S = 15 m to the door of the carriage, the train started to move

When the passenger had to travel the distance S = 15 m to the door of the carriage, the train started to move and began to accelerate with constant acceleration, the modulus of which was a = 0.5 m / s ^ 2. The passenger began to catch up with the train, moving at a constant speed, the modulus of which is U = 4 m / s. After what is the minimum period of time it will reach the carriage door.

The passenger reached the carriage door:

S2 = S + S1.

S2 is the distance covered by the passenger (S2 = V * t, where V is the speed (V = 4 m / s), t is the time).

S – initial distance to the door (S = 15 m).

S1 is the distance that the train will travel (S1 = V0 + at ^ 2/2, where V0 = 0 m / s, a is the acceleration (a = 0.5 m / s2)).

V * t = S + at ^ 2/2.

4t = 15 + 0.25t ^ 2.

0.25t ^ 2 – 4t + 15 = 0 | * 4.

t ^ 2 – 16t + 60 = 0.

According to Vieta’s theorem:

t1 + t2 = -p, where p = -16.

t1 * t2 = q, where q = 60.

Roots: t1 = 6 s, t2 = 10 s.