Kinematics Ques 60
- A particle is moving with a velocity $\mathbf{v}=k(y \hat{\mathbf{i}}+x \hat{\mathbf{j}})$, where $k$ is a constant.
The general equation for its path is
(2019 Main, 9 Jan I)
(a) $y=x^{2}+$ constant
(b) $y^{2}=x+$ constant
(c) $x y=$ constant
(d) $y^{2}=x^{2}+$ constant
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Answer:
Correct Answer: 60.(d)
Solution:
Formula:
- Given, velocity of a particle is
$$ \mathbf{v}=k(y \hat{\mathbf{i}}+x \hat{\mathbf{j}}) $$
Suppose, it’s position is given as
$$ \begin{array}{rlrl} & \mathbf{r} & =x \hat{\mathbf{i}}+y \hat{\mathbf{j}} \\ \therefore & \mathbf{v} & =\frac{d \underline{\mathbf{r}}}{d t} & =\frac{d}{d t}(x \hat{\mathbf{i}}+y \hat{\mathbf{j}})=\frac{d x}{d t} \hat{\mathbf{i}}+\frac{d y}{d t} \hat{\mathbf{j}} \end{array} $$
Comparing Eqs. (i) and (ii), we get
and
$$ \begin{aligned} & \frac{d x}{d t}=y \\ & \frac{d y}{d t}=x \end{aligned} $$
Dividing Eq. (iii) and Eq. (iv), we get
$$ \frac{\frac{d x}{d t}}{\frac{d y}{d t}}=\frac{y}{x} \Rightarrow x \frac{d x}{d t}=y \frac{d y}{d t} \quad \text { or } x d x=y d y $$
Integrating both sides, we get
$$ \int x d x=\int y d y \text { or } \frac{x^{2}}{2}+\frac{c _1}{2}=\frac{y^{2}}{2}+\frac{c _2}{2} $$
where, $c _1$ and $c _2$ are the constants of integration.
$$ \begin{aligned} & \Rightarrow \quad x^{2}+c=y^{2} \quad\left[\text { here }, c(\text { constant })=c _1-c _2\right] \\ & \text { or } \quad y^{2}=x^{2}+\text { constant } \end{aligned} $$
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