A particle moves according to v² = u² + 2αx - βx², where α,β>0. Considering all physically accessible positions, which statements are correct?
Select all that apply

Question

A particle moves according to v² = u² + 2αx − βx², where α,β>0. Considering all physically accessible positions, which statements are correct?

G1 Physics · Accepted Answer

Differentiate the velocity-position relation: 2v dv/dx = 2α−2βx. Since a = v dv/dx, we obtain a = α−βx. Thus A is correct. Speed is maximum where d(v²)/dx=0, 2α−2βx=0 ⇒ x=α/β. Therefore B is correct. At x=α/β, vₘₐₓ² = u² + 2α(α/β) − β(α²/β²) = u² + α²/β. So C is correct. Turning positions satisfy v=0: βx² − 2αx − u² = 0. Hence x = [α ± √(α²+βu²)]/β. The distance between the two roots is 2√(α²+βu²)/β. Therefore D is also correct. All four statements are correct.