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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

The displacement x (in metre) of a particle of mass m kg moving in one dimension under the action of a force is related to time t (in second) by the equation t = √x + 3. The work done by the force in the first six seconds is

Answer: (B)

Solution

x = (t−3)², so v = dx/dt = 2(t−3). At t=0, v=−6; at t=6, v=+6. Since |v| is the

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Question ID VK-153721