The initial speed of train is v0=90km/h=25m/s and the acceleration is −0.5m/s2. According to the kinematic equation:
v0=−at where t is the time required to stop the train. Thus:
t=−av0=−−0.525=50s The distance, traveled by train by this time is:
d=v0t+2at2=25⋅50−20.5⋅502=625m Answer. d = 625 m.
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