Answer:
  dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))
Explanation:
- Equation of motion for constant acceleration are:
                    v = a*t
                   z =( a*t^2)/2 Â
                   v^2 = 2*z*s
- Compute time t1:
                  x = (a*t^2)/2
                  t_1 = sqrt(2*x/a)
- Velocity @ time t is: v = a*t = a*sqrt(2*x/a)
- Remaining distance (s-x) is run at above calculated speed, so the time t_2 is:
               t_2 = (s-x)/v = (s-x)/(a*sqrt(2*x/a))
- Compute total T:
          T = t_1 + t_2 = sqrt(2*x/a) + (s-x)/(a*sqrt(2*x/a))
- Simplify: Â Â Â Â Â Â Â Â Â T = (x+s)/sqrt(2*a*x)
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- Time taken by each runner:
                 T_A =(x+s)/sqrt(2*a1*x)
                T_B = (x+s)/sqrt(2*a2*x)+ dt
Equating the total times (T_A = T_B): Â
           (x+s)/sqrt(2*a2*x)+ dt = (x+s)/sqrt(2*a1*x)
Hence,
          dt = (x+s)/sqrt(2*a1*x) - (x+s)/sqrt(2*a2*x))
          dt = (x+s)/sqrt(2x)*(1/sqrt(a1) - 1/sqrt(a2))