Hey!

Differentiate t with respect to x,

= dt/dx = d(ax² + bx)/dx

dt/dx = 2ax + b .....(1)

from equation (1), (v = dx/dt)

dt/dx = 1/{dx/dt} = 1/v = 2ax + b

v = 1/(2ax + b) ......(2)

now differentiating v with with respect to time, t, we get acceleration =a=

dv/dt = d{1/(2ax + b)}/dt

= -1/(2ax + b)² × d(2ax + b)/dt

= -1/(2ax + b)² × [2a × dx/dt ]

= -1/(2ax + b)² × 2a v

from equation (2),

dv/dt = -v² × 2av = -2av³

Which means retardation is - 2av³

[here negative sign shows retardation.]

Thank you.