Since it is braking, the acceleration is negative.
Use the standard equation
Vf^2-Vi^2=2(a)S
where
Vf = final velocity (=0)
Vi = initial velocity (90 km/h=90*1000/3600=25 m/s)
a=acceleration required (m/s^2)
S=distance = 1.3 km = 1300 m
0^2-25^2=2(a)1300
Solve for a (in m/s^2)
a train stopping distance, even when full emergency brakes are engaged, is 1.3 km. if the train was travelling at an initial velocity of 90 km/h [forward], determine its acceleration under full emergency brakes.
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