/*
   Lab : Mock Test 01
   Day : Tuesday  [ 16 Sept 08 ]

   Problem Number: 02
   Problem Title : Minimum Travel Time of a Sub-Way Train
   
   Theory :
	when train starts with initial velocity u=0 and reaches to maximum velocity v=maxVelocity, with constand acceleration a=maxAcceleration.
        the time taken to reach maximum velocity can be computed by the formula v = u + a*t

        i.e.  t = maxVelocity / maxAcceleration

	in achieving maximum speed the distance traveled by the train can be calculated by s = u*t + 0.5*a*t^2

	i.e. 	s = 0*t + 0.5*a*t^2
		s = 0.5 * maxAcceleration * t * t
		s = 0.5 * maxAcceleration * ( maxVelocity / maxAcceleration ) * ( maxVelocity / maxAcceleration )
		s = 0.5 * ( maxVelocity * maxVelocity ) / maxAcceleration

	The train journey has three parts

	1. s distance with acceleration a in time t
	2. ( distance -2s ) with constant speed in time ( distance -2s ) /  maxVelocity
	3. s distance with retardation a in time t

	Total travel time is sum of all three

  
    Important Note:
      It is also possible that the distance between two stations is not enough to speed up the train up to 
      its maximum speed. 

      Q: When this will happen ? 
      A: Time taken to attain maximum speed by train
            t = maxVelocity / maxAcceleration  
         distance traveled in this duration (use s = u*t + 0.5*a*t*t)
            ( 0.5  * maxVelocity * maxVelocity / maxAcceleration  ) 
         
         if the distance between two consecutive station is less than 
               2 * ( 0.5  * maxVelocity * maxVelocity / maxAcceleration ) 
            that is 
               ( maxVelocity * maxVelocity / maxAcceleration )
         then train can not attain its maximum speed .


         for this case 
         the train starts from speed 0 accelerates with speed maxAcceleration until 
         it reaches to the half of distance b/w two stations
         and then retards with same acceleration maxAcceleration until it stops.

         
      Q: What is time taken to go up to half of distance b/w two stations
      A: using formula  s = u*t + 0.5*a*t*t;
         as u = 0  the formula becomes   s = 0.5*a*t*t;    
	   or as     t = Squire root of ( 2*s/a )

         here s=distance/2;       a=maxAcceleration;     
         therefore    
				t = Squire root of ( 2*(distance/2) / maxAcceleration ) 
				t = Squire root of ( distance / maxAcceleration ) 

         In Java we can use function Math.sqrt for squire root of therefore 
                        t = Math.sqrt( distance / maxAcceleration )
 
         Since this is time to cover half the distance, therefore time to cover full distance
                          2 * Math.sqrt( distance / maxAcceleration )
 */

class sol_MockTest01_prob02{

	
   	public static void main(String args[])
	{		
		int distance, maxVelocity, maxAcceleration;
		double s, t_acceleration, t_constantSpeed, t_total;
		
		distance        = Integer.parseInt(args[0]) ;
		maxVelocity     = Integer.parseInt(args[1]) ;
		maxAcceleration = Integer.parseInt(args[2]) ;

		// First check whether the distance b/w two stations 
		// is not sufficient to attain maximum speed by train
            	if(  ( maxVelocity * maxVelocity / maxAcceleration ) > distance )
		{			
			t_total = 2 * Math.sqrt( distance / maxAcceleration );
		}		
		else
		{       // Time to accelerate
			t_acceleration = ( double ) maxVelocity / maxAcceleration ;
			s = ( 0.5 *  maxVelocity * maxVelocity )  / maxAcceleration ;

			// Time of constant journy
			t_constantSpeed = ( double ) ( distance - 2*s ) /  maxVelocity ;

			// keeping in mind acceleration and retardation time are same 
			t_total = t_acceleration + t_constantSpeed + t_acceleration  ;
		}
		
		System.out.println(" Minimum journey time is "+ t_total + " sec"); 	

	}	
}