A train is travelling at a speed of 90 km/h the breaks are applied so as to produce a uniform acceleration of -0.5m/s^2.find how far the train goes before it stops?
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v =0
u=90km/h=25m/s
a=-0.5m/s^2
second equation of motion
=v^2-u^2=2as
=0-625=2×(-0.5)×s
=-625=-s
s=625m
u=90km/h=25m/s
a=-0.5m/s^2
second equation of motion
=v^2-u^2=2as
=0-625=2×(-0.5)×s
=-625=-s
s=625m
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