A particle is moving with speed v = bsqrt{x} | vedclass

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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

Vedclass · Accepted Answer

$\frac{b^2	au}{2}$

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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 $t = 	au$ (where $x = x$):$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{	au} b dt$$[2\sqrt{x}]_{0}^{x} = b	au$$2\sqrt{x} = b	au$,which implies $\sqrt{x} = \frac{b	au}{2}$.Substituting this into the expression for speed $v = b\sqrt{x}$:$v = b \left( \frac{b	au}{2} ight) = \frac{b^2	au}{2}$.

$A$ 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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