A car starting from rest accelerates at the rate f through a distance s1 then continues at constant speed for time t and then decelerate at the rate f/2 to come to rest. If the total distance travelled is 15s then s is

As the car velocity is decreasing with time We can divide this question in 3 steps  1. When car moving with acceleration f and covers the distance s Initial speed =u= 0  Final velocity after moving a distance ‘S’ is vequation of motion ,   v2−u2=2as ,  substitute,v2=2fs⇒v=2fs---(1)here s1=s    2. In seconds part the car moving with constant speed hence a=0   Then distance=s2=ut+12at2 ⇒s2=2fst---(2)   3. After this car is decelerated by f/2  The distance =s3final velocity v2=0v22−u2=2as30−2fs=−2f/2s3⇒s3=2s---(3)  The total distance travel is 15s=s1+s2+s315s=s+2fst+2s12s=2fst∴s=ft272