A point initially at rest moves along the x axis. Its acceleration varies with time as $ a = (6t + 5)\,m\,{s^{ - 2}} $ . If it starts from origin, the distance covered in 2s is
(A) 20m
(B) 18m
(C) 16m
(D) 25m

Hint :In this question we shall use the concept of instantaneous acceleration and instantaneous velocity at a given time t. Given the instantaneous parameters, we can find the total parameters under a given time interval by integrating it over a certain specified range. Since in the question, the particle starts from the origin and no initial time is mentioned, we shall take the initial conditions to be $ x = 0\,,\,t = 0 $ .

Complete Step By Step Answer:
We know that the acceleration is the derivative of the velocity with respect to time which is in turn a derivative of distance with respect to time. So, the relationship between the acceleration and the distance covered can be established as
 $ a = \dfrac{{dv}}{{dt}} $ where v is the velocity of the particle.
 $ \int {adt} = \int {dv} $
 $ \Rightarrow \int {adt} = v $
Given that the acceleration of the particle varies with time as $ a = (6t + 5)\,m\,{s^{ - 2}} $ .
Integrating once we get,
 $ \int {(6t + 5)dt} = v $
 $ \Rightarrow \dfrac{{6{t^2}}}{2} + 5t = v $
 $ \Rightarrow 3{t^2} + 5t = v $
Now we know that $ v = \dfrac{{dx}}{{dt}} $
 $ \Rightarrow \int {vdt} = x $
Given that the velocity of the particle varies with time as $ v = (3{t^2} + 5t)\,m\,{s^{ - 1}} $ .
Integrating once we get,
 $ \int {(3{t^2} + 5t)dt} = x $
 $ \Rightarrow \dfrac{{3{t^3}}}{3} + \dfrac{{5{t^2}}}{2} = x $
 $ \Rightarrow {t^3} + \dfrac{{5{t^2}}}{2} = x $
At $ x = 2 $ ,
 $ x = {2^3} + \dfrac{{5 \times {2^2}}}{2} $
 $ x = 18\,m $
Hence option B is the correct answer.

Note :
The actual expression for the calculation of the integral should be $ \int_{{t_1}}^{{t_2}} {vdt = x} $ . This formula is a general expression and eliminates the constant also. But since the initial was not specified, we took $ {t_1} = 0 $ . Thus, we didn’t have to apply limits over the integral and simply substituting $ x = 2 $ gave us the correct answer.