A train 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.

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07-01-2017
Physics

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A train 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.

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MissPhiladelphia MissPhiladelphia · Answer 1 · 2017-01-11T08:25:48-05:00

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