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JEE Main 2005 (AIEEE) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2005 (AIEEE)

Q37JEE Main 2005 · AIEEEDifficult

Equations of Kinematics

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 f2\frac{f}{2} to come to rest. If the total distance traversed is 15 S, then

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Answer: (d)

  1. After distance SS: v2=2fSv^2 = 2fS.
  2. Deceleration at f2\frac{f}{2}: distance =v22(f/2)=2fSf=2S= \frac{v^2}{2(f/2)} = \frac{2fS}{f} = 2S.
  3. Constant-speed part: vt=15S−S−2S=12S⇒v=12Stvt = 15S - S - 2S = 12S \Rightarrow v = \frac{12S}{t}
  4. 144S2t2=2fS⇒S=172ft2\frac{144S^2}{t^2} = 2fS \Rightarrow S = \frac{1}{72}ft^2
  5. So the answer is (d) S=172ft2S = \frac{1}{72}ft^2.
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