Question

The motion of a particle along a straight line is described by equation x=8+12t-t3 where x is in metre and t in second. The retardation of the particle when its velocity become zero is

Answer 1

retardation=12m/s2
refer to the image attached for solution

Answer 2

The particle’s retardation is $12\ m/s^2$

Given:

$x=8+12 t-t^{3}$

Solution:

We know that,

On differentiating, displacement, we get velocity.

$\Rightarrow \text { Velocity }(v)=\frac{d x}{d t}=12-3 t^{2}$

Again on differentiating, velocity, we get acceleration.

$\Rightarrow \text { Acceleration }(a)=\frac{d v}{d t}=-6 t$

When velocity becomes zero,

$12-3 t^{2}=0 \Rightarrow 12=3 t^{2} \Rightarrow 4=t^{2} \Rightarrow t=2 s$

Therefore,

$=-6 t=-6 \times 2=-12 \ \mathrm{m} / \mathrm{s}^{2}$

In the above solution, the answer we got is negative, that’s because it is negative acceleration, also known as retardation.

To calculate the retardation, we took the approach of calculus and we used the concept that the acceleration is the differentiation of velocity which in turn is derived by differentiating the distance. By definition, acceleration is rate of change of velocity.