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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeHard2 minQB

PhysicsHardsingle choice

On a two-lane road a car $A$ travels at $2 m / s$ . Cars $B$ and $C$ approach $A$ from opposite directions at $10 m / s$ each. When $B$ and $C$ are each $1200$ m from $A$ , $B$ decides to overtake $A$ before meeting $C$ . The minimum acceleration of $B$ required is:

Options:

Answer:

Solution:

Relative to $A$ : $B$ moves at $8 m / s$ and $C$ at $12 m / s$ , so $C$ reaches $A$ in $\frac{1200}{12} = 100$ s. For $B$ : $1200 = 8 (100) + \frac{1}{2} a 10 0^{2}$ , giving $a = \frac{400}{5000} = 0.08 m / s^{2}$ . Option B.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 2mℹ️ Source: QB

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