Question

The acceleration $\alpha$ of a particle moving with an velocity $u$ varies with distance $x$ as $a=k\sqrt{x}$ where $k$ is a constant . Th distance covered by the particle when its velocity become $3u$ is given by

• A. ${\left(\frac{3u^2}{k}\right)}^{2/3}$
• B. ${\left(3k{u}^2\right)}^{4/3}$
• C. ${\left(\frac{6u^2}{k}\right)}^{2/3}$
• D. ${\left(\frac{u^2}{3k}\right)}^{2/3}$

Answer 1

The correct option is C ${\left(\frac{6u^2}{k}\right)}^{2/3}$

Since, we have given

$a=k\sqrt{2}$

we know that,

$a=\frac{vdv}{dx}$

$\Rightarrow v\frac{dv}{dx}=k\sqrt{x}$

$\Rightarrow {\int }_u^{3u}vdv={\int }_0^{x}k{x}^{\frac{1}{2}}dx$

$\Rightarrow \frac{1}{2}[v^2{]}_u^{3u}=\frac{2}{3}k[x^{3/2}{]}_0^x$

$\Rightarrow \frac{1}{2}[9u^2-u^2]=\frac{2}{3}k-x^{3/2}$

$\Rightarrow \frac{1}{2}\times 84=\frac{2}{3}k{x}^{3/2}$

$\Rightarrow x^{3/2}=\frac{6u^2}{k}$

$\Rightarrow x={\left(\frac{{64}^2}{k}\right)}^{2/3}$.