Atrain travels a distance of 1,2 km between two stations with an average velocity of 43.2 km/h. during it's motion, at the time t1=40s it moved accelerated, then at time t2 it moved uniformly, then at t3=40s it moved uniformly slowed. find the maximum velocity of the train. the acceleration during the slowed motion is equal in modulus to the acceleration during the accelerated motion.

For the answer to the question above,
The whole trip covered d = 1.2 km at 43.2 km/hr taking a total time of T = (1.2 / 43.2) hr = 100 s. Since T = t1 + t2 + t3 and t1=t3=40 s, then t2 = 100-40-40 = 20 s. 

The area under the v vs. t curve is the distance d traveled. The shape of the curve is an isosceles trapezoid (trapezium in UK English) with bases T and t2 and height V. So, from the area formula for a trapezoid: 

d = (V/2)(T + t2) 
V = 2D / (T + t2) = (2.4 km) / (120 s) = 0.02 km/s 

Multiply by 3600 s/hr to get 72 km/h to match the original units in the problem

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