A particle moves in the x-y plane with a velocity, $\overrightarrow{v}=2y\widehat{i}+4\widehat{j}$ . Equation of the path followed by the particle is:
$\begin{align}
& (A)x=\sqrt{y} \\
& (B)y=\sqrt{x} \\
& (C)4x={{y}^{2}} \\
& (D)4y={{x}^{2}} \\
\end{align}$
Answer
552.3k+ views
Hint: We will first find the X and Y-components of the velocity using the given velocity equation in the problem. Once we get these terms, they can be written as a differential of displacement with respect to time in X and Y directions respectively. After this, we shall proceed to create a differential equation in X and Y so as to get our required path that the particle is following.
Complete answer:
The velocity vector has been given as:
$\Rightarrow \overrightarrow{v}=2y\widehat{i}+4\widehat{j}$
Now, let the X-component of the velocity vector be denoted by $\overrightarrow{{{v}_{x}}}$ .
And, let the Y-component of the velocity vector be denoted by $\overrightarrow{{{v}_{y}}}$ .
Then, these will be equal to:
$\Rightarrow {{v}_{x}}=2y$ And,
$\Rightarrow {{v}_{y}}=4$
Now these components of velocity in the X and Y direction could be written as the differentials of the displacement with respect to time in the respective directions. That is:
$\Rightarrow \dfrac{dx}{dt}=2y$ [Let this expression be equation number (1)]
$\Rightarrow \dfrac{dy}{dt}=4$ [Let this expression be equation number (2)]
On dividing equation number (2) by (1), we get:
$\begin{align}
& \Rightarrow \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{4}{2y} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{4}{2y} \\
& \Rightarrow 2y(dy)=4(dx) \\
\end{align}$
Thus, we get a linear differential in X and Y. This could be easily solved as follows:
Integrating both sides of the equation, we get:
$\Rightarrow \int{2y.dy}=\int{4.dx}$
$\Rightarrow 2\left[ \dfrac{{{y}^{2}}}{2} \right]=4\left[ x \right]$ $\left[ \text{Using: }\int{{{a}^{n}}dn=\dfrac{{{a}^{n+1}}}{n+1}} \right]$
$\therefore {{y}^{2}}=4ax$
Hence, the equation of the path followed by the particle is given by ${{y}^{2}}=4ax$.
Hence, option (C) is the correct option.
Note:
In this problem we saw the use of basic definition of velocity and how we used it to calculate the path of the particle. Also, whenever dealing with questions like these we should have a basic knowledge of integral and differential calculus as they too played a key part in solving the problem. Formulas for all the basic integrals should be remembered thoroughly.
Complete answer:
The velocity vector has been given as:
$\Rightarrow \overrightarrow{v}=2y\widehat{i}+4\widehat{j}$
Now, let the X-component of the velocity vector be denoted by $\overrightarrow{{{v}_{x}}}$ .
And, let the Y-component of the velocity vector be denoted by $\overrightarrow{{{v}_{y}}}$ .
Then, these will be equal to:
$\Rightarrow {{v}_{x}}=2y$ And,
$\Rightarrow {{v}_{y}}=4$
Now these components of velocity in the X and Y direction could be written as the differentials of the displacement with respect to time in the respective directions. That is:
$\Rightarrow \dfrac{dx}{dt}=2y$ [Let this expression be equation number (1)]
$\Rightarrow \dfrac{dy}{dt}=4$ [Let this expression be equation number (2)]
On dividing equation number (2) by (1), we get:
$\begin{align}
& \Rightarrow \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{4}{2y} \\
& \Rightarrow \dfrac{dy}{dx}=\dfrac{4}{2y} \\
& \Rightarrow 2y(dy)=4(dx) \\
\end{align}$
Thus, we get a linear differential in X and Y. This could be easily solved as follows:
Integrating both sides of the equation, we get:
$\Rightarrow \int{2y.dy}=\int{4.dx}$
$\Rightarrow 2\left[ \dfrac{{{y}^{2}}}{2} \right]=4\left[ x \right]$ $\left[ \text{Using: }\int{{{a}^{n}}dn=\dfrac{{{a}^{n+1}}}{n+1}} \right]$
$\therefore {{y}^{2}}=4ax$
Hence, the equation of the path followed by the particle is given by ${{y}^{2}}=4ax$.
Hence, option (C) is the correct option.
Note:
In this problem we saw the use of basic definition of velocity and how we used it to calculate the path of the particle. Also, whenever dealing with questions like these we should have a basic knowledge of integral and differential calculus as they too played a key part in solving the problem. Formulas for all the basic integrals should be remembered thoroughly.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

