- Home
- Standard 11
- Physics
3-2.Motion in Plane
hard
A particle is moving with velocity $\vec v = K(y\hat i + x\hat j)$ where $K$ is a constant. The general equation for its path is
A
$y^2 = x^2 + $ constant
B
$y = x^2 + $ constant
C
$y^2 = x +$ constant
D
$xy =$ constant
(AIEEE-2010) (JEE MAIN-2019)
Solution
$\begin{array}{l}
\vec v = k\left( {y\hat i + x\,\hat j} \right) = {v_x}\hat i + {v_y}\hat j = \frac{{dx}}{{dt}}\hat i + \frac{{dy}}{{dt}}\hat j\\
\therefore \frac{{dx}}{{dt}} = ky\,\,\,\,\,and\,\,\,\,\therefore \frac{{dy}}{{dt}} = kx\\
\therefore \frac{{dy}}{{dx}} = \frac{x}{y} \Rightarrow ydy = xdx \Rightarrow {y^2} = {x^2} + {\rm{constant}}
\end{array}$
Standard 11
Physics
