Me out to solve this problem because it looks difficult

the time when he starts to deaccelerate will be =

distance between a and b at the starting = x-y = 12.6m

Explanation:

writing equation of motion for traveller A v = u + at

where v is the final  u the initial velocity and a is the acceleration

u = 10 v= 8 and a = -5 applying the values in the equation we get t =

therefore athlete A deaccelerates from 10 to 8 m/s in s

so the time when he starts to deaccelerate will be = 2 -

                                                                                    =

writing equation of motion for B

v = u + at

where v= 8 u = 0 and t = 2

therefore applying the values a= 4

therefore distance travelled be B in 2 =

therefore   =x+ 8m

for traveller b distance travelled in 2  = y + 10× + 10× + -5×

= y+19.6 m

given

therefore x-y-11.6 = 1

therefore distance between a and b at the starting = x-y = 12.6m