Question

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

Answer 1

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.