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JEE Main 2023 (1 Feb, Shift 2) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2023 (1 Feb, Shift 2)

Q18JEE Main 2023 · 1 FEB · SHIFT 2Medium

Equations of Kinematics

For a train engine moving with speed of 20 ms−120\text{ ms}^{-1}, the driver must apply brakes at a distance of 500 m\text{m} before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed x ms−1\sqrt{x}\text{ ms}^{-1}. The value of xx is ____________.

(Assuming same retardation is produced by brakes)

Numerical answer type. Work it out, then open the answer.

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Answer: 200

  1. Retardation: 0=202−2a(500)⇒a=0.4 m/s20 = 20^2 - 2a(500) \Rightarrow a = 0.4\text{ m/s}^2.
  2. Braking over 250 m: v2=202−2(0.4)(250)=400−200=200v^2 = 20^2 - 2(0.4)(250) = 400 - 200 = 200
  3. v=200 m/sv = \sqrt{200}\text{ m/s}, so x=200x = 200
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