Question

A particle is moving with velocity $v=k\left(y\hat{\text{i}}+x\hat{\text{j}}\right)$ where $k$ is a constant. The general equation for its path is

• A. $y=x^2+constant$
• B. $y^2=x+constant$
• C. $xy=constant$
• D. $y^2=x^2+constant$

Answer 1

Answer: (D)

$v=k\left(y\hat{\text{i}}+x\hat{\text{j}}\right)$
$\frac{\text{d}\text{x}}{\text{d}\text{t}}=\text{ky}$
$\frac{\text{dy}}{\text{d}\text{t}}=\text{kx}$
$\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{\text{d}\text{y}}{\text{d}\text{t}}\cdot \frac{\text{d}\text{t}}{\text{d}\text{x}}=\frac{\left(\text{kx}\right)}{\left(\text{ky}\right)}$
$\left(\text{d}\text{y}\right)\text{ky}=\text{d}\text{x.kx}$
$\text{y}\text{dy}=\text{x}\text{d}\text{x}$
$\frac{{\text{y}}^2}{2}=\frac{{\text{x}}^2}{2}+\text{constant}$
$\Rightarrow {\text{y}}^2={\text{x}}^2+\text{constant}$