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 f/2 to come to rest. If the total distance  traversed is 15 s, then

See figureGiven that s + s1 + s2 = 15 s ...(1) For motion along OA, we have v2=2fs(∵u=0 and v=v)For motion along AB, s1 = v t …(2)For motion along BCv2=2(f/2)s2....(3)          (∵u=v,v=0 and a=−f/2)Solving above equations, we haves=172ft2