The train starts from ground a at rest and accelerates to ground B at a constant acceleration of a, then decelerates to ground C at a constant acceleration of a / 3 in the original direction and stops If the distance between a and C is 28.8km, the vehicle has been running for 24min

According to s = V * t + 1 / 2 * a * T ^ 2, suppose that the distance between a and B is S1, and the time is T1. The distance between B and C is S2, and the time is T2, then: S1 = 0.5 * a * T1 ^ 2s2 = a * T1 * t2-0.5 * A / 3 * T2 ^ 2 because S1 + S2 = 28.8, T1 + T2 = 24, so: 0.5 * a * T1 ^ 2 + A * T1 * (24-t1) - 0.5 * A / 3 * (24-t1) ^ 2 = 28.8 -------- 1

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