Q77 / 90 · JEE Main 2019 (12 Apr Shift 2) · Physics

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

A
$b^{2} τ$
B
$\frac{b^{2} τ}{\sqrt{2}}$
Correct answer
C
$\frac{b^{2} τ}{2}$
D
$\frac{b^{2} τ}{4}$

Solution

v = b \sqrt{x}

… (1)

\Rightarrow \frac{d x}{d t} = b \sqrt{x}

\frac{d x}{\sqrt{x}} = b d t

\int_{0}^{x} \frac{d x}{\sqrt{x}} = \int_{0}^{t} b d t

2 \sqrt{x} = b t

… (2)
From equation (1) and (2)

v = b (\frac{b t}{2})

∴ a t t = τ

v = \frac{b^{2} τ}{2}

Hence, the correct answer is option: 'C'.

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