Question

a train starts from rest from a station with acceleration 0.2 metre per second on a straight track and then comes to rest after attaining maximum distance on another station due to reterdation 0.4 M per second .If Total time is 30 min find distance between two station

Answer 1

There are direct equations for such type of questions,

α = positive acceleration; β = Retardation;T= total time; S = Total diatance.
So train moves with positive acceleration α till attaining max velocity and then retards with β, in time T.
So, α=0.2m/s^2;β= 0.4m/s^2 ; T= 30 min= 1800sec
S = αβT∧2÷2(α+β)
Therefore, S= [0.2×0.4 ×( 1800)^2]/ [ 2(0.2+0.4)] = 216000 m = 216 km