Question

The distance $x$ covered by a particle in one dimension motion varies as with time as $x^2=a{t}^2+2bt+c$, where $a,b,c$ are constants. Acceleration of particle depend on $x$ as $x^{-n}$, the value of $n$ is;

Answer 1

The explanation for the correct answer.

1. Step 1: Given Data:

The equation of motion is $x^2=a{t}^2+2bt+c$

Velocity $V=\frac{dx}{dt}$and acceleration $\alpha =\frac{dV}{dt}$

Step 2: Calculation:

$\begin{array}{rcl}{x}^2& =& a{t}^2+2bt+c\\ 2xv& =& 2at+2b\\ xv& =& at+b\dots \dots \dots \dots \left(i\right)\\ v& =& \frac{(at+b)}{x}\end{array}$

Now, differentiating equation (i)

$\begin{array}{rcl}{v}^2+x& =& a\\ \alpha x& =& a–\frac{{(at+b)}^2}{x^2}\\ \alpha & =& a(a{t}^2+2bt+c)-\frac{(at+b^2)}{x^3}\\ \alpha & =& ac–\frac{b^2}{x^3}\\ \alpha & \propto & x^{-3}\end{array}$

Thus, the value of $n$ is $3.$