Question

The acceleration of a body moving with initial velocity $u$ changes with distance $x$ as $a=k^2\sqrt{x}$where $k$ is a positive constant. The distance travelled by the body when its velocity becomes $2u$ is

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

Answer 1

The correct option is B

${\left(\frac{3u}{2k}\right)}^{\frac{4}{3}}$

Step 1: Given data

$a=k^2\sqrt{x}$

Step 2: To find

The distance travelled by the body when its velocity becomes $2u$.

Here, $a$ is the acceleration, $x$ is the distance, $u$ is the initial velocity, $v$ is the final velocity, $k$ is a positive constant and $t$ is the time taken.

Step 3: Formula and calculation

We know,

$$\begin{gathered} a=\frac{dv}{dx}\times \frac{dx}{dt} \\ a=v\frac{dv}{dx} \end{gathered}$$

$$\begin{gathered} \therefore v\frac{dv}{dx}=k^2\sqrt{x} \\ \Rightarrow vdv=k^2\sqrt{x}dx \end{gathered}$$

Integrating,

$$\begin{gathered} \begin{array}{l}{\int }_u^{2u}vdv={\int }_0^x{k}^2\sqrt{x}dx\end{array} \\ \Rightarrow {\left[\frac{v^2}{2}\right]}_u^{2u}={\left[\frac{2k^2{x}^{{\displaystyle \frac{3}{2}}}}{3}\right]}_0^x \\ \Rightarrow \frac{3u^2}{2}=\frac{2k^2{x}^{\frac{3}{2}}}{3} \\ \Rightarrow x={\left(\frac{3u}{2k}\right)}^{\frac{4}{3}} \end{gathered}$$

Therefore, the distance travelled when its velocity becomes $2u$ is ${\left(\frac{3u}{2k}\right)}^{\frac{4}{3}}$.