Question

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

Answer 1

v = 44.27 m / s

Step-by-step 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