A train is stopped by putting a fixed brake), so it stops at 50 meters away. If the train has already doubled it and stopped the train with the same deficit force, then how far will it stop to stop
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Answer:
200 m
Explanation:
Let the velocity when brake was applied is u m/s.
final velocity, v=0 m/s
retardation, a= -a m/s²
stopping distance, s= 50 m
On applying IIIrd equation of motion,
v²=u²+2as
0=u² - 2a(50)
100a = u²
a = u²/100............. (1)
Now speed of the train is doubled i.e., it becomes 2u m/s.
final velocity, v=0
retardation, a = - u²/100 m/s² [From eq1]
stopping distance, s = s metres
On applying IIIrd equation of motion,
v²=u²+2as
0=(2u)² - (2u²s) /100
(u²s)/50 = 4u²
s = 200 m
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