Two cyclists are riding towards each other. One, having a speed of 10 km / h, moves with equal slowness

| June 19, 2021 | education

Two cyclists are riding towards each other. One, having a speed of 10 km / h, moves with equal slowness with an acceleration of 0.2 m / s2, the other, having a speed of 5.4 km / h, moves uniformly with an acceleration of 0.2 m / s2. How long will the cyclists meet and what movement will each of them make before the meeting, if the distance between them at the initial moment of time is 130 m?

We translate the speed from km / h to m / s: 10 km / h = 25/9 m / s, 5.4 km / h = 1.5 m / s.

The first cyclist rode S1 = 25/9 * t – 0.1 * t ^ 2.

The second cyclist rode S2 = 1.5 * t + 0.1 * t ^ 2.

Together they drove: S1 + S2 = 130,

25/9 * t – 0.1 * t ^ 2 + 1.5 * t + 0.1 * t ^ 2 = 130,

77/18 * t = 130,

t = ~ 30.39 sec = ~ 0.5 min (but the first cyclist will already be standing by this time)

S1 = V0 ^ 2/2 * a = (25/9) ^ 2 / (2 * 0.2) = ~ 19,

S2 = 130 – S1 = ~ 111.



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