If the maximum acceleration that is tolerable for passengers in a subway train is 1.21 m/s2 and subway stations are located 810 m apart, what is the maximum speed a subway train can attain between stations

v = 44.27 m / s

Explanation:

This is an exercise in kinematics since they ask the maximum speed of the train

           v² = v₀² + 2 a x

a train departs with zero initial velocity

           v² = 2 a x

           v = √ 2ax

calculate

           v = √ (2 1.21 810)

            v = 44.27 m / s

in this solution we have assumed that the train uses the entire distance with acceleration

ice to water 1 water to vapour 2 latent heat

no

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