Question

A particle moves according to the law $a=-ky$. Find the velocity as a function of distance $y,v_0$ is initial velocity and particle is at origin initially.

• A. $v^2=v_0^2-k{y}^2$
• B. $v^2=v_0^2-2ky$
• C. $v^2=v_0^2-2k{y}^2$
• D. None

Answer 1

The correct option is A $v^2=v_0^2-k{y}^2$

$a=\frac{dv}{dt}=\frac{dv}{dy}.\frac{dy}{dt}$
where $v=\frac{dy}{dt}$
$\underset{v_0}{\overset{v}{\int }}vdv=\underset{0}{\overset{y}{\int }}-kydy$ or $v_0^2-v^2=k{y}^2$