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The distance $x$ covered by a particle varies with time $t$ as ${{x}^{2}}=2{{t}^{2}}+6t+1$. Its acceleration varies with x as:
$\begin{align}
  & A.\,\,x \\
 & B.\,\,{{x}^{2}} \\
 & C.\,\,{{x}^{-1}} \\
 & D.\,\,{{x}^{-3}} \\
 & E.\,\,{{x}^{-2}} \\
\end{align}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: To get the acceleration of a particle, we need to double differentiate the distance with respect to time, which will give us the value of acceleration of the particle. After first differentiation, we will get the velocity of the particle and when we again differentiate the obtained equation, we will get the acceleration of the particle with respect to time.

Complete Step-by-Step solution:
It is given that the distance $x$ covered by a particle varies with time $t$ as ${{x}^{2}}=2{{t}^{2}}+6t+1$.

Now,When we differentiate the above equation with respect to $t$, we will get the velocity of the particle.

So, after differentiating the particle with respect to $t$, we get:
 $2x\dfrac{dx}{dt}=4t+6$
In the above equation, $\dfrac{dx}{dt}=v$, i.e., velocity, then
$\begin{align}
  & 2xv=4t+6 \\
 & \Rightarrow xv=2t+3 \\
\end{align}$

Again, differentiating the above equation with respect to $t$, we get:
 $x\dfrac{dv}{dt}+v\dfrac{dx}{dt}=2$

In the above equation, $\dfrac{dx}{dt}=v$, i.e., velocity and $\dfrac{dv}{dt}=a$, i.e., acceleration, then
$xa+{{v}^{2}}=2$

Now, substituting the value of $v$ in the above equation from the equation $xv=2t+3$, we get:
$\begin{align}
  & xa+{{\left( \dfrac{2t+3}{x} \right)}^{2}}=2 \\
 & \Rightarrow a{{x}^{3}}+4{{t}^{2}}+12t+9=2{{x}^{2}} \\
 & \Rightarrow a{{x}^{3}}+2(2{{t}^{2}}+6t+1)+7=2{{x}^{2}} \\
\end{align}$

We know that, ${{x}^{2}}=2{{t}^{2}}+6t+1$.
Then,
$\begin{align}
  & a{{x}^{3}}+2({{x}^{2}})+7=2{{x}^{2}} \\
 & \Rightarrow a{{x}^{3}}+7=0 \\
 & \therefore a=\dfrac{-7}{{{x}^{3}}} \\
\end{align}$

Hence acceleration varies with ${{x}^{-3}}$.

Therefore, the correct answer is Option (D).

Additional Information:
In a compressible sound transmission medium - mainly air - air particles get an accelerated motion: the particle acceleration or sound acceleration with the symbol a in metre/second2. In acoustics or physics, acceleration (symbol: a) is defined as the rate of change (or time derivative) of velocity. It is thus a vector quantity with dimension length/time2. In SI units, this is $dfrac{m}{s^2}$.

Note:
This is a simple question of differentiation. We need to keep one thing in mind that what we will get after differentiating the equation. Just remember that when we differentiate distance, we get velocity and when we differentiate velocity, we get acceleration, this is a very important relationship to remember.