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`        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.`
10 years ago

```							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
S2 = 1350 mts frm decelerating point
So at distances  225 mts and 2250 mts ,starting from rest , it has half max. speed

```
10 years ago
```							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!!!!
```
10 years ago
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