A particle is moving with speed v = b √(x) along positive x-axis. Calculate the speed of the particle at time t = τ (assume that the particle is at origin t = 0) (a) (b²τ)/(√(2)) (b) b²τ (c) (b²τ)/(2) (d) (b²τ)/(4)

Answer: (c). (dx)/(√(x)) = b dt ⇒ 2√(x) = bt (since x = 0 at t = 0) x = (b² t²)/(4) v = dx/dt = (b² t)/(2); at t = τ, v = (b²τ)/(2) So the answer is (c).

A particle is moving with speed v = b √(x) along positive x-axis… | JEE Main 2019 (12 Apr, Shift 2)

Q31JEE Main 2019 · 12 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion

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

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Answer: (c)

$\dfrac{dx}{\sqrt{x}} = b\,dt \Rightarrow 2\sqrt{x} = bt$ (since $x = 0$ at $t = 0$ )
$x = \dfrac{b^2 t^2}{4}$
$v = \dfrac{dx}{dt} = \dfrac{b^2 t}{2}$ ; at $t = \tau$ , $v = \dfrac{b^2\tau}{2}$
So the answer is (c).

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Velocity, Acceleration & Variable Motion question from JEE Main 2019 (12 Apr, Shift 2)

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