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

Question

Arun · Answer

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

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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

S = 1/6 ft²

S = ft

S = ¼ ft²

S = 1/72 ft²

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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 S = 1/6 ft2 S = ft S = ¼ ft2 S = 1/72 ft2

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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 ft²

S = ft

S = ¼ ft²

S = 1/72 ft²