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\tS = 1/6 ft2S = ftS = ¼ ft2S = 1/72 ft2
Anantha vishnu R , 5 Years ago
Grade 12
1 Answers
Arun
Last Activity: 5 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
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