Question

A particle moves in the XY-plane according to the law $x=kt,y=kt(1-\alpha t)$ where $k$ and $\alpha$ are positive constants and $t$ is time. The trajectory of the particle is:

• A. $y=kx$
• B. $y=x-\frac{\alpha x^2}{k}$
• C. $y=-\frac{\alpha x^2}{k}$
• D. $y=\alpha x$

Answer 1

The correct option is B $y=x-\frac{\alpha x^2}{k}$

The given co-ordinates are $x=kt$ $y=kt(1-\alpha t)$

We get $t=\frac{x}{k}$

Putting this in the equation of y co-ordinates.

We get $y=k\times \frac{x}{k}(1-\alpha \times \frac{x}{k})$

or $y=x(1-\frac{\alpha x}{k})$

$⟹y=x-\frac{\alpha x^2}{k}$