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# A particle moves in a straight line and its position x at time t is given by ${{x}^{2}}=2+t$. Its acceleration is given by$\text{A}\text{. }\dfrac{-2}{{{x}^{3}}}$$\text{B}\text{. }\dfrac{-1}{4{{x}^{3}}}$$\text{C}\text{. }\dfrac{-1}{4{{x}^{2}}}$$\text{D}\text{. }\dfrac{1}{{{x}^{2}}}$

Last updated date: 18th Jun 2024
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Hint: The second derivative of the position of the particle with respect to time is called the acceleration of the particle. Hence, differentiate the given equation i.e. ${{x}^{2}}=2+t$ twice to find the second derivative of x with respect to t. i.e. $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$.

Let us first understand what is meant by acceleration of the particle. Before that we have to understand the meaning of velocity of a particle.
Velocity of the particle is defined as the rate of change of position of the particle with respect to time. We can also say the velocity of the particle is the change in position of the particle in one unit of time.
When we say that a particle is in an acceleration, it means its velocity is changing with time. Acceleration of a body is defined as the rate of change of velocity of a particle with respect to time. In other words, it is the change in velocity of the moving particle in one unit of time.
Mathematically, acceleration is the second derivative of the position of the moving particle with respect to time i.e.$\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$. Therefore, if we want to find the acceleration of a particle, all we want is the function of its position with respect to time i.e. x(t). Then the second derivative of the function x(t) with respect to time t will give us the acceleration of the particle.
In the question, the square of the position of the particle as a function of time t is given as ${{x}^{2}}=2+t$ ….. (i).
Differentiate both the sides of equation (i) with respect to x.
Therefore, we get
$2x\dfrac{dx}{dt}=1$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{1}{2x}$…. (ii)
Therefore, we found the first derivative of x with respect to time (i.e. velocity of the particle).
Now, differentiate equation (ii) with respect to x.
Hence, we get,
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{-1}{2{{x}^{2}}}\dfrac{dx}{dt}$.
Substitute the value of $\dfrac{dx}{dt}$ from equation (ii).
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{-1}{2{{x}^{2}}}.\dfrac{1}{2x}=\dfrac{-1}{4{{x}^{3}}}$
This means that the second derivative of the position of the particle is $\dfrac{-1}{4{{x}^{3}}}$. Hence, the acceleration of the particle is $\dfrac{-1}{4{{x}^{3}}}$.

So, the correct answer is “Option B”.

Note: We can also first write the function of x by taking square root on both the sides of equation (i) and differentiate x twice with respect to time t to find the acceleration. When we take square root, x(t) becomes $x=\pm \sqrt{2+t}$.
However, students may make a mistake by just taking the positive value of the root i.e. $x=\sqrt{2+t}$ and neglect the negative value. Note that both the values must be taken into consideration.