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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2019 · 12 April · Evening Shift · Physics · Motion in One Dimension

A particle is moving with speed v = b x along positive x - axis....

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 = τ (assume that the particle is at origin at t=0t = 0 )

Options

  1. Ab2τ{b}^{2}τ
  2. Bb2τ2\dfrac{{b}^{2}τ}{\sqrt{2}}
  3. Cb2τ2\dfrac{{b}^{2}τ}{2}
  4. Db2τ4\dfrac{{b}^{2}τ}{4}

Correct answer

C. b2τ2\dfrac{{b}^{2}τ}{2}

Step-by-step solution

v=bxv = b\sqrt{x} … (1)
⇒dxdt=bx⇒ \dfrac{dx}{dt} = b\sqrt{x}
dxx=bdt\dfrac{dx}{\sqrt{x}} = bdt
∫0xdxx=∫0tbdt\mathop{∫}_{0}^{x}\dfrac{dx}{\sqrt{x}} = \mathop{∫}_{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 = τ
v=b2τ2v = \dfrac{{b}^{2}τ}{2}
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