Question

A particle moves according to the law a = -ky. Find the velocity as a function of distance y, v₀ is initial velocity.

• A. v² = v₀² – ky²
• B. v² = v₀² – 2ky
• C. v² = v₀² – 2ky²
• D. None

Answer 1

Answer: (A)

v² = v₀² – ky²

Answer 2

a = $\frac{dv}{dt} = \frac{dv}{dy}.\frac{dy}{dt}$

or $\int_{v_{0}}^{v}{vdv} = \int_{0}^{y}{- kydy}$ or v₀² – v² = ky².