Question #284852

  1. A bullet train accelerates from rest at Tokyo Station to 89 m/s over a distance of 120 m. What acceleration did it experience?

Expert's answer

The acceleration is


a=v2/(2d)=892/(2240)=16.5 m/s2.a=v^2/(2d)=89^2/(2·240)=16.5\text{ m/s}^2.


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