Question

A particle starts moving at $t = 0$ from origin along the (+ve) direction of x-axis. Its speed depends on the distance travelled $x$ as $v = K\sqrt{x}$. If $v = \frac{dx}{dt}$ and $a = \frac{dv}{dt}$, then find $x$, $v$ and $a$ in terms of time $t$.

• A. $\frac{Kt^2}{4}$, $\frac{Kt}{2}$, $\frac{K}{2}$
• B. $\frac{K^2t^2}{4}$, $\frac{K^2t}{2}$, $\frac{K^2}{2}$
• C. $\frac{Kt^2}{4}$, $\frac{Kt^2}{2}$, $\frac{K^2}{2}$
• D. $\frac{K^2t^2}{4}$, $\frac{Kt}{2}$, $\frac{K^2}{2}$

Answer 1

Answer: (B)

$\frac{K^2t^2}{4}$, $\frac{K^2t}{2}$, $\frac{K^2}{2}$

Answer 2

We are given:

$$\frac{dx}{dt}=K\sqrt{x}.$$

Separate variables:

$$\frac{dx}{\sqrt{x}} = K\, dt.$$

Integrate:

$$\int \frac{dx}{\sqrt{x}} = \int K\, dt \quad\Longrightarrow\quad 2\sqrt{x} = Kt.$$

Thus,

$$\sqrt{x} = \frac{Kt}{2}\quad\Longrightarrow\quad x = \frac{K^2t^2}{4}.$$

Find velocity $v$:

$$v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{K^2t^2}{4}\right)=\frac{K^2t}{2}.$$

Also, verifying with $v=K\sqrt{x}$:

$$v=K\left(\frac{Kt}{2}\right)=\frac{K^2t}{2}.$$

Find acceleration $a$:

$$a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{K^2t}{2}\right)= \frac{K^2}{2}.$$

Thus, we have:

$$x=\frac{K^2t^2}{4},\quad v=\frac{K^2t}{2},\quad a=\frac{K^2}{2}.$$