KinematicsHard
Question
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 traversed is 15 S, then
Options
A.S = ft
B.S = 1/6 ft2
C.S = 1/2 ft2
D.S = 1/4 ft2
Solution
S =
v0 =
During retardation
S2 = 2S
During constant velocity
15S - 3S = 12S = vot
⇒ S = ft2/2
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