The relationship between time $ t $ and distance $ x $ is defined as $ t = ax^2 + bx $, where $ a $ and $ b $ are constants.

To find the acceleration, follow these steps:

First, differentiate the equation with respect to time $ t $:

$ \frac{d}{dt}(t) = a \frac{d}{dt}(x^2) + b \frac{dx}{dt} $

This simplifies to:

$ 1 = a \cdot 2x \frac{dx}{dt} + b \frac{dx}{dt} $

Given $ \frac{dx}{dt} = v $ (where $ v $ is the velocity), we can write:

$ 1 = 2axv + bv = v(2ax + b) $

This simplifies to:

$ 2ax + b = \frac{1}{v} $

Next, differentiate the expression $ 2ax + b = \frac{1}{v} $ again with respect to $ t $:

$ 2a \frac{dx}{dt} + 0 = -\frac{1}{v^2} \frac{dv}{dt} $

Using $ \frac{dx}{dt} = v $, we get:

$ 2av = -\frac{1}{v^2} \frac{dv}{dt} $

Solving for $ \frac{dv}{dt} $:

$ \frac{dv}{dt} = -2a v^3 $

Since acceleration $ f = \frac{dv}{dt} $, we have:

$ f = -2av^3 $