(B) Given the velocity $v = \frac{dx}{dt} = \alpha \sqrt{x}$.Rearranging the terms to separate the variables,we get:$\frac{dx}{\sqrt{x}} = \alpha dt$.

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(B) Given the velocity $v = \frac{dx}{dt} = \alpha \sqrt{x}$.Rearranging the terms to separate the variables,we get:$\frac{dx}{\sqrt{x}} = \alpha dt$.Integrating both sides with the initial condition $x=0$ at $t=0$:$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \alpha dt$.Solving the integral:$[2x^{1/2}]_{0}^{x} = \alpha [t]_{0}^{t}$.This simplifies to:$2\sqrt{x} = \alpha t$.Squaring both sides:$4x = \alpha^2 t^2$.Therefore,the displacement $x$ is proportional to $t^2$:$x \propto t^2$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

Medium

$A$ particle located at $x= 0$ at time $t= 0$ starts moving along the positive $x-$direction with a velocity $v$ that varies as $v = \alpha \sqrt{x}$. The displacement of the particle varies with time as:

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A
$t^3$
B
$t^2$
C
$t$
D
$t^{1/2}$

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Motion in Straight Line

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Uniformly Accelerated Motion

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$A$ particle located at $x= 0$ at time $t= 0$ starts moving along the positive $x-$direction with a velocity $v$ that varies as $v = \alpha \sqrt{x}$. The displacement of the particle varies with time as:

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