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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 ft 2 S = ft S = ¼ ft 2 S = 1/72 ft 2

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
  1. S = 1/6 ft2       
  2. S = ft 
  3. S = ¼ ft2
  4.  S = 1/72 ft2

Grade:12

1 Answers

Arun
25763 Points
2 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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