A train starts from rest and moves with a constant acceleration of 2.0 m/s^2 for half a minute

We solve it in two parts, 1.accelerated motion and 2. decelerated motion.  

For 1.accelerated motion  

u=0, a=2.0 m/s². t=½*60 =30 seconds, 

Now final velocity is given by v=u+at =0+2*30 =60 m/s  

Since after this velocity brakes are applied, so it is the  

(b) maximum speed attained = 60 m/s. 

Distance traveled x=ut+½at² =0+½*2*30² =900 m =0.9 km  

Let the distance traveled at half the maximum speed (30 m/s) be y, 

Now from the equation v²=u²+2ax, 

30²=0²+2*2*y →y=900/4 =225 m .........(1)

 For decelerated motion   

u=60 m/s, v=0 m/s, t=60 s From the equation v=u+at  

a=(0-60)/60 = -1 m/s²  

Distance  traveled = ut+½at² =60*60+½*-1*60² = 3600-1800 =1800 m =1.8 km  

Hence answer for (a) The total distance moved = 0.9 km + 1.8 km  =2.7 km    

Let the distance moved at half the speed in this part be z   

From v²=u²+2ax →30²=60²+2*-1*z → 2z=60²-30²=2700 m →z=1350 m =1.35 km ....(2)

 (c) There will be two positions at half the maximum speed (From the point of start),      

 From ..(1) =225 m and From ...(2) 1.35 km+0.9 km = 2.25 km