Question

The velocity where $v$of a particle as a function of its position (x) is expressed as$v=\sqrt{c_1-c_{2}x},$
$c_1$ 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

Answer: (B)

$-\frac{c_2}{2}$

Given,

$v=\sqrt{c_1-c_{2}x}$

Acceleration, $a=v\frac{dv}{dt}$

Differentiate $v$ with respect to $x$

$\frac{dv}{dx}=\frac{1}{2\sqrt{c_1-c_{2}x}}\times (-c_2)$

$\frac{dv}{dx}=-\frac{c_2}{2v}$

$v\frac{dv}{dx}=-\frac{c_2}{2}$

$a=-\frac{c_2}{2}$

The correct option is B.