The motion of an object along a straight line is described by the equation: $x=10+14t-t^3$. Here x is in metres and t in seconds. What is the particle's retardation when its velocity reaches zero?

The rate of change in velocity with respect to time is measured by acceleration, which can be positive or negative in value.

$a=\frac{v-u}{t}$

Here v= Final velocity, u= Initial velocity and t= time

Given

 $-x=10+14t-t^3$

The rate of change of displacement is what we call velocity.

$v=\frac{dx}{dt}$

    $=\frac{d(10+14t-t^3)}{dt}$

     $=14-3t^2$ 

if v=0, then

$14-3t^2=0$

14=3t²

$t=2.16 sec$

The retardation is given by

$a=\frac{-dv}{dt}$

    $=\frac{-d(14-3t^2)}{dt}$

     $=-(-6t)$

      $=6\times 2.16$

      =$13 m{s}^{-2}$