# A car, starting from rest, accelerates at the rate ‘f’ through a distance S , then continues at constant speed for time t and then deccelerates at the rate f/2 to come to rest. If the total distance covered is 15S, then S = 1/6 ft2        S = ft  S = ¼ ft2  S = 1/72 ft2

Arun
25758 Points
3 years ago
the car covers the total distance in three phases.
distance in phase 1=s........velocity(v) after this phase:v2=2fs
distance in phase 2:s2=vt=?(2fs)  x  t
distance in phase 3: s3=putting in third eq of motion:0=v2-2xf/2 xs3
s3=2s...putting the value of v2
now,   s+?(2fs)  x  t+2s=15s
solving this eq: s=2ft2/72