Kinematics

JEE Main level, single correct, medium

133 of 308

Question 133 of 308

A particle is moving with speed v=bxv=b\sqrt{x} along positive x−x- axis. Calculate the speed of the particle at time t=τt=\tau (assume that the particle is at origin at t=0t=0 )

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Answer: (C) b2τ2\dfrac{{b}^{2}\tau }{2}

v=bxv=b\sqrt{x} … (1)
⇒dxdt=bx\Rightarrow \dfrac{dx}{dt}=b\sqrt{x}
dxx=bdt\dfrac{dx}{\sqrt{x}}=bdt
∫0xdxx=∫0tbdt\int _{0}^{x}\dfrac{dx}{\sqrt{x}}=\int _{0}^{t}bdt
2x=bt2\sqrt{x}=bt … (2)
From equation (1) and (2)
v=b(bt2)v=b\left(\dfrac{bt}{2}\right)
∴att=τ∴att=\tau
v=b2τ2v=\dfrac{{b}^{2}\tau }{2}

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