Q#36 (Past Exam Paper – June 2013 Paper 12 Q7)
The diagram shows an arrangement to stop trains that are travelling too fast.
Trains coming from the left travel at a speed of 50 m s-1. At marker 1, the driver must apply the brakes so that the train decelerates uniformly in order to pass marker 2 at no more than 10 m s-1.
The train carries a detector that notes the times when the train passes each marker and will apply an emergency brake if the time between passing marker 1 and marker 2 is less than 20 s.
How far from marker 2 should marker 1 be placed?
A 200 m B 400 m C 500 m D 600 m
Solution:
Answer: D.
Initial speed u = 50 m s-1
Final speed (max) v = 10 m s-1
Minimum time for passing between marker 1 and 2: t = 20 s
{If time is less than 20 s, it means that the train is moving too fast. So an emergency brake is then applied.}
Since the train undergoes uniform deceleration, we can use the equations of uniformly accelerated motion.
Distance between markers = s = ???
s = ut + ½ at2
To obtain the acceleration a:
a = (v-u) / t = (10 – 50) / 20 = - 2 m s-2
Replacing a in the equation,
s = ut + ½at2
s = (50×20) + (½×-2×20)2
s = 600 m
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