Question

For a given motion, the relationship between time t and distance x is found out to be t=ax^2+bx, where a and b are constants. Considering v as velocity, the retardation will be given by,

Answer 1

$$t=ax^2+bx$$

Differentiate both side with respect to time:

$$1=2ax\frac{dx}{dt}+ b\frac{dx}{dt}=2axv+bv$$

$$v^{-1}=2ax+b$$

Differentiate both side with respect to time:

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

Th retardation is

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