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JEE Main 2019 · 12 Apr · Shift II

JEE Main 2019 Physics question on Kinematics

A particle is moving with speed v=bxv = b\sqrt{x} along positive XX-axis. Calculate the speed of the particle at time t=τt = \tau (assume that the particle is at origin at t=0t = 0).
  1. A.
    b2τ4\dfrac{b^2\tau}{4}
  2. B.
    b2τ2\dfrac{b^2\tau}{2}
  3. C.
    b2τb^2\tau
  4. D.
    b2τ2\dfrac{b^2\tau}{\sqrt{2}}
Show answer

Correct option: B

[JEE Main 2019 · 12 Apr · Shift II]

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