Question

The distance $x$ covered by a particle in one-dimensional motion varies with time $t$as $x^2=a{t}^2+2bt+c$. The acceleration of the particle varies as

• A. $x^{-3}$
• B. $x^{\frac{3}{2}}$
• C. $x^2$
• D. $x^{\frac{-2}{3}}$

Answer 1

The correct option is A

$x^{-3}$

Find the relation of acceleration

$x^2=a{t}^2+2bt+c$

Let $v$ be the velocity and $a_c$be the acceleration

Differentiating the given equation with respect to $t$

$$\begin{gathered} \Rightarrow \frac{d{x}^2}{dt}=\frac{d(a{t}^2+2bt+c)}{dt} \\ \Rightarrow 2xv=2at+2b \\ \Rightarrow v=\frac{at+b}{x} \end{gathered}$$

Differentiating again with respect to $t$

$\Rightarrow v^2+a_{c}x=a$

Putting the value of $v$ here

$$\begin{gathered} \Rightarrow {\left(\frac{at+b}{x}\right)}^2+a_{c}x=a \\ \Rightarrow a_c\alpha x^{-3} \end{gathered}$$

Hence, the correct option is $A$.