Question

If ( mathrm{A} ) is the acceleration of the train and ( mathrm{B} )
is the retardation of the train, then the
total distance travelled is given by ( [(mathrm{AB}) /(mathrm{A}+mathrm{B})] cdotleft(mathrm{T}^{wedge} 2 / 2right. )
Here ( A=0.2 mathrm{m} / mathrm{s}^{wedge} 2 ) and ( mathrm{B}=0.4 mathrm{m} / mathrm{s}^{wedge} 2 )
and ( T=1 / 2 mathrm{hr}=1800 mathrm{sec} )
Therefore ( x=[(0.2) )
( (0.4) / 0.2+0.4)] times 1800 times 1800 / 2=216000 m )
( =216 mathrm{km} )

# Awo coste costwo 5 varies o are article w Board & CompellIVO 26. À train starts from rest from a station with acceleration 0.2 m/s2 on a straight track and then comes to rest after attaining maximum speed on another station due to retardation 0.4 m/s2. If total time spent is half an hour, then distance between two stations is [Neglect length of train] (1) 216 km (3) 728 km (2) 512 km (4) 1296 km

Solution