Both the cars are accelerating towards right and their initial velocity is also in rightward direction as shown in figure. Maximum separation between car A and car B (in meters) is $50 α$ . The value of $α$ is

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v A ​ = v B ​ . 10 + 2 t = 20 + t ⇒ t = 10 s s A ​ = 10 ( 10 ) + 2 1 ​ ( 2 ) ( 1 0 2 ) = 200 m s B ​ = 20 ( 10 ) + 2 1 ​ ( 1 ) ( 1 0 2 ) = 250 m Δ s = 250 − 200 = 50 m 100 + 50 = 150 m 150 m = 50 α , then α = 3 .

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JEE Main - 1 year (12th class)