Answer
438k+ views
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. $\dfrac{{{d}^{2}}x}{d{{t}^{2}}}$.
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)