A train starts from rest and moves with constant acc. of 2 m/s^2 for half a minute. The brakes are then apllied and it comes to rest in 1 minute. Find a) Total distance moved. b)max. speed attained by train and c) the position(s) of train at half the max. speed.

the 3 relations used here are : v=u+at          --- 1

                                                        S= ut + at²/2  --- 2

                                                        v² - u² = 2aS   --- 3     u = initial vel

                                                                                      v= final velocity and a = acceleration  S = distance travelled

For first 1/2 min,  u =0 and a₁ =2 and t₁ = 30 sec

so final vel.    v₁ = at₁        ( using relation 1)

  v₁=60 m/sec

after applying brakes, the train starts decelerating

so here v₁=u₂= 60 m/sec and t₂= 60 sec and a₂ = -a₁              ( using relation 1)

so (a₂) = 1 m/sec²

so total distance covered = 1/2*a₁*t₁² + u₂² / 2*a₂  
                                               = 1/2*2*30² + 60² / 2*1                 ( using relation 2 and 3)

                                         a) so total distance covered     =   2700 m

                               b)max. speed attained by train is v₁=60 m/sec

                               c) the position(s) of train at half the max. speed.

to find positions at speed of 30 m/sec

during accln. : v=u+at                   30 = 2 t   so t = 15 sec

S = 1/2*2*15² whihc gives 225 mts

S₁ =225 mts

during deceleration :  v=u+at                30 = 60 -t          so t=30 sec

v² - u² = 2aS                so  60² - 30² = 2S₂

S_{2 =} 1350 mts frm decelerating point

So at distances  225 mts and 2250 mts ,starting from rest , it has half max. speed

try solvig this question using graphs.

draw the velocity time graph of the motion of train starting from zero velocity and finally comming back to zero.

slope during acceleration is 2.

max speed=0+2*30=60m/s

it reaches half the max speed at 15secs.

total displacement is equal to the area under the graph.

positon of train at half the max speed is the area under the graph till the half max speed.

i think its the shortest solution and also the quickest!!!!