A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f2 to come to rest. If the total distance traversed is 15 S, then

Let car starts from point A from rest and moves up to point B with acceleration f.Velocity of car at point B, v =2fS                                                                [As v2= u2+2as]x = 2fS .t  −−−(i)        [As s = ut}So the velocity of car at point C also will be 2fs and finally car stops after covering distance y.Distance CD i.e, y = (2fS)22(f2) = 2fSf = 2S  ...(ii)                               [As v2= u2−2as  s = u2/2a]So, total distance AD = AB +BC +CD = 15S(given)⇒ S+x+2S = 15S  ⇒ x = 12SSubstituting the value of x in equation (i) we getX =  2fS.t ⇒ 12S = 2fS .t⇒ 144S2 = 2fS.t2⇒ S = 172ft2