The relation between time ‘ t ’ and distance ‘ x ’ is t = α x ^ 2 +...

Given: $t = \alpha{x}^{2} + \betax$ 

Differentiating w.r.t time, we get

$\dfrac{dt}{dx} = 2\alphax + \beta$

$⇒ \dfrac{1}{v} = 2\alphax + \beta$

Differentiating w.r.t time again, we get

$⇒ - \dfrac{1}{{v}^{2}}\dfrac{dv}{dt} = 2\alpha\dfrac{dx}{dt}$

$⇒ \dfrac{dv}{dt} = - 2\alpha{v}^{3}\left[\text{as}v = \dfrac{dx}{dt}\right] ⇒ a = \dfrac{dv}{dt} = - 2\alpha{v}^{3}$