The relation between the’t’ and distance 'x' is t = ax2+bx, where a & b are constants. The  retardation is:  (a) 2a.v3 (b) 2b.v3 (c) 2ab.v3 (d) 2b2v3

The given relation is t = ax^2 + bx.

First, we need to find the velocity 'v'. Velocity is the rate of change of distance with respect to time. So, we differentiate 'x' with respect to 't' to get the velocity.

dx/dt = v = 2ax + b

Next, we need to find the retardation or deceleration. Retardation is the rate of decrease of velocity, which is the negative of acceleration. Acceleration is the rate of change of velocity with respect to time. So, we differentiate 'v' with respect to 't' to get the acceleration.

dv/dt = a = 2a

Since retardation is the negative of acceleration, the retardation is -2a.

However, none of the given options match this result. There might be a mistake in the question or the options provided.