Question

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

NEET/Medical Exams

Physics

Solution

115

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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} )