A particle is moving along a straight line path according to the relation :-
${{s}^{2}}=a{{t}^{2}}+2bt+c$
Where s represents the distance travelled in t seconds and a,b and c are constants. Then the acceleration of the particle is given as
A. $\dfrac{a}{s}-\dfrac{{{(at+b)}^{2}}}{{{s}^{3}}}$
B. $\dfrac{a}{s}-\dfrac{{{(at+b)}^{2}}}{{{s}^{2}}}$
C. $\dfrac{at}{s}-\dfrac{{{(at+b)}^{2}}}{{{s}^{3}}}$
D. $\dfrac{at}{s}-\dfrac{{{(at+b)}^{2}}}{{{s}^{2}}}$

Hint:In this question, a particle is moving along a straight line path and we have to find out the acceleration of the particle. Acceleration is the rate of change of velocity. We find out the acceleration of the particle by differentiating the given equation two times. By differentiating it two times, we get the desirable answer.

Formula used:
Acceleration is defined as the a = \[\dfrac{{{d}^{2}}s}{d{{t}^{2}}}\]
Where s represents the distance travelled in t seconds and a,b,c are constants.

Complete step by step solution:
Given a particle is moving along a straight line according to the equation:-
${{s}^{2}}=a{{t}^{2}}+2bt+c$………………………………….. (1)
We have to find the acceleration of the particle. Acceleration can be defined as a vector quantity that defines the rate at which an object changes its velocity. We know that,
\[a = \dfrac{{{d}^{2}}s}{d{{t}^{2}}}\]

Differentiating equation (1) w.r.t t, we get
$2s\dfrac{ds}{dt}=2ab+2b$
$\Rightarrow \dfrac{ds}{dt}=\dfrac{at+b}{s}$
We again differentiate the above equation, we get
$\dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\dfrac{a.s-(at+b)\dfrac{ds}{dt}}{{{s}^{2}}} \\ $
Now we put the value of $\dfrac{ds}{dt}$in the above equation, we get
\[\dfrac{{{d}^{2}}s}{d{{t}^{2}}}\]= \[\dfrac{as-(at+b)\left( \dfrac{at+b}{s} \right)}{{{s}^{2}}} \\ \]
$\Rightarrow \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\dfrac{a{{s}^{2}}-{{(at+b)}^{2}}}{{{s}^{3}}}$
Solving further, we get
$\therefore \dfrac{{{d}^{2}}s}{d{{t}^{2}}}=\dfrac{a}{s}-\dfrac{{{(at+b)}^{2}}}{{{s}^{3}}}$

Thus, option A is correct.

Note: Acceleration is used to describe the motion of the body and determine the force. It is a vector quantity and can be translational and rotational. Students must be careful in differentiating the terms.