Question

The velocity $v$ of a particle as a function of its position $\left(x\right)$ is expressed as $v=\sqrt{c_1-c_{2}x}$, where $c_1$ and $c_2$ are positive constants. The acceleration of the particle is

• A. $c_2$
• B. $-\frac{c_2}{2}$
• C. $c_1-c_2$
• D. $\frac{c_1+c_2}{2}$

Answer 1

The correct option is B $-\frac{c_2}{2}$

Acceleration $=a=\frac{dv}{dt}\text{or}a=v\frac{dv}{dx}$
$v=\sqrt{c_1-c_{2}x}$
$\Rightarrow v^2=c_1-c_{2}x\dots \dots \dots \left(i\right)$
Differentiating $\left(i\right)$ w.r.t to $x$, we get
$2v\frac{dv}{dx}=-c_2$
$\therefore a=v\frac{dv}{dx}=-\frac{c_2}{2}$