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JEE Main 2019 · 9 Jan · Shift II

JEE Main 2019 Physics question on Kinematics

In a car race on a straight path, car AA takes a time tt less than car BB at the finish and passes finishing point with a speed 'vv' more than that of car BB. Both the cars start from rest and travel with constant acceleration a1a_1 and a2a_2 respectively. Then 'vv' is equal to
  1. A.
    2a1a2a1+a2t\dfrac{2a_1 a_2}{a_1 + a_2} t
  2. B.
    2a1a2 t\sqrt{2a_1 a_2}\, t
  3. C.
    a1a2 t\sqrt{a_1 a_2}\, t
  4. D.
    a1+a22t\dfrac{a_1 + a_2}{2} t
Show answer

Correct option: C

[JEE Main 2019 · 9 Jan · Shift II]

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