Between points A and B the distance is 4000m. A car drove out of point A at a constant speed of 20m / s.

Between points A and B the distance is 4000m. A car drove out of point A at a constant speed of 20m / s. After 50 s, the second car left the same point at a speed of 30 m / s. How far from point B will the second car catch up with the first?

S = 4000 m.

V1 = 20 m / s.

V2 = 30 m / s.

t = 50 s.

Sв -?

Before the meeting, the first car will travel the distance S1, which we express by the formula: S1 = V1 * t1, where V1 is the speed of the first car, t1 is the time the first car is moving before the meeting.

Before the meeting, the second car will travel the distance S2, which we will express by the formula: S2 = V2 * t2, where V2 is the speed of the second car, t2 is the time of movement of the second car before the meeting.

S2 = S1.

t1 = t2 + t.

V1 * t1 = V2 * t2.

V1 * (t2 + t) = V2 * t2.

V1 * t2 + V1 * t = V2 * t2.

V2 * t2 – V1 * t2 = V1 * t.

t2 = V1 * t / (V2 – V1).

t2 = 20 m / s * 50 s / (30 m / s – 20 m / s) = 100 s.

S2 = 30 m / s * 100 s = 3000 m – the second car will meet the first from point A.

Sv = S – S2.

Sw = 4000 m – 3000 m = 1000 m.

Answer: at a distance of Sв = 1000 m from point B, the second car will catch up with the first.



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