Mon June 11, 2012 By: Shubhankar

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 travelled is 15S, then S

Expert Reply
Mon June 11, 2012
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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