Question

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

• 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 of the above

Answer 1

Answer: (A)

$v^2={v_0}^2-k{y}^2$

$a=-ky\Rightarrow v\frac{dv}{dy}=-ky\Rightarrow vdv=-kydy\Rightarrow {\int }_{v_o}^{v}vdv={\int }_0^y-kydy\Rightarrow \frac{v^2-v_o^2}{2}=-\frac{k{y}^2}{2}\Rightarrow v^2=v_o^2-k{y}^2$