Question

The speed $v$ of a particle moving along a straight line is given by $a+b{v}^2=x^2$ (where $x$ is its distance from the origin). The acceleration of the particle is

• A. $bx$
• B. $\frac{x}{a}$
• C. $\frac{x}{b}$
• D. $\frac{x}{ab}$

Answer 1

The correct option is C

$\frac{x}{b}$

Step 1: Given Data:

Equation of motion, $a+b{v}^2=x^2$

where $v$ is the speed of particle moving along a straight line, and $x$ is its distance from the origin.

Step 2: Formula used:

The rate at which velocity changes with time, is known as acceleration.

The rate at which displacement changes with time is known as velocity.

$\begin{array}{rcl}v& =& \frac{dx}{dt}\dots \dots \dots \dots \dots \dots \left(1\right)\\ a& =& \frac{dv}{dt}\dots \dots \dots \dots \dots \dots \left(2\right)\end{array}$

where $v,t,x,\mathrm{and}a$are velocity, time, distance, and acceleration respectively.

Step 3: Calculation of acceleration of a particle:

The given equation of motion is $a+b{v}^2=x^2\dots \dots \dots \dots \dots \dots \dots \left(3\right)$

Differentiating the given equation of motion (3) with respect to $t$. We get,

$\begin{array}{rcl}0+2bv\frac{dv}{dt}& =& 2x\left(\frac{dx}{dt}\right)\\ bv\frac{dv}{dt}& =& x\left(\frac{dx}{dt}\right)\end{array}$

Substituting the values from equation (1) $v=\frac{dx}{dt}$and from equation (2), $a=\frac{dv}{dt}$

$\begin{array}{rcl}bv\frac{dv}{dt}& =& x\frac{dx}{dt}\\ bv\frac{dv}{dt}& =& xv\\ \frac{dv}{dt}& =& \frac{x}{b}\\ a& =& \frac{x}{b}\end{array}$

Hence option C is the correct answer.