Question

A car accelerates from rest at a constant rate α a for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is :

(1) \(\frac{4\alpha\beta}{(\alpha+\beta)}t^2\)

(2) \(\frac{2\alpha\beta}{(\alpha+\beta)}t^2\)

(3) \(\frac{\alpha\beta}{2(\alpha+\beta)}t^2\)

(4) \(\frac{\alpha\beta}{4(\alpha+\beta)}t^2\)

Answer 1

Correct option is  (3) \(\frac{\alpha\beta}{2(\alpha+\beta)}t^2\)

v₀ = αt₁ and 0 = v₀ – βt₂

⇒ v₀ = βt₂

t₁ + t₂ = t

Distance = area of v - t graph