(B) Given the speed $v = b\sqrt{x}$.We know that $v = \frac{dx}{dt}$,so $\frac{dx}{dt} = b\sqrt{x}$.Separating the variables,we get $\frac{dx}{\sqrt{x

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(B) Given the speed $v = b\sqrt{x}$.We know that $v = \frac{dx}{dt}$,so $\frac{dx}{dt} = b\sqrt{x}$.Separating the variables,we get $\frac{dx}{\sqrt{x}} = b dt$.Integrating both sides from $t = 0$ (where $x = 0$) to $t = \tau$ (where $x = x$):$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{\tau} b dt$$[2\sqrt{x}]_{0}^{x} = b\tau$$2\sqrt{x} = b\tau$,which implies $\sqrt{x} = \frac{b\tau}{2}$.Substituting this into the expression for speed $v = b\sqrt{x}$:$v = b \left( \frac{b\tau}{2} \right) = \frac{b^2\tau}{2}$.

Class 11 Physics Motion in Straight Line Uniformly Accelerated Motion

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

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A
$b^2\tau$
B
$\frac{b^2\tau}{2}$
C
$\frac{b^2\tau}{\sqrt{2}}$
D
$\frac{b^2\tau}{4}$

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Motion in Straight Line

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Uniformly Accelerated Motion

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

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$A$ body is moving with a uniform velocity of $8\,m/s$. When the body just crosses another body,the second body starts from rest and moves with a uniform acceleration of $4\,m/s^2$. The time after which the two bodies meet will be $...........\,s$.

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