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JEE Main 2024 (9 Apr, Shift 2) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2024 (9 Apr, Shift 2)

Q7JEE Main 2024 · 9 APR · SHIFT 2Easy

Equations of Kinematics

Two cars are travelling towards each other at speed of 20 m s−120\text{ m s}^{-1} each. When the cars are 300 m300\text{ m} apart, both the drivers apply brakes and the cars retard at the rate of 2 m s−22\text{ m s}^{-2}. The distance between them when they come to rest is:

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Answer: (b)

  1. Stopping distance of each car: s=u22a=2022×2=100 ms = \frac{u^2}{2a} = \frac{20^2}{2 \times 2} = 100\text{ m}.
  2. Both cars together cover 2×100=200 m2 \times 100 = 200\text{ m}.
  3. Gap left =300−200=100 m= 300 - 200 = 100\text{ m}.
  4. So the answer is (b) 100 m.
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