If the velocity as a function of position is given as $v = 2x$ .Find $v$ at $t = 2\sec $,if it’s given that $x = 2$ at $t = 0$:
A. $4{e^4}$
B. $2ln4$
C. $6m{\sec ^{ - 1}}$
D. $2{e^4}$

Hint: In order to solve this question, we need to convert given velocity-position relation into a velocity-time dependent function and then will find the value of velocity at time $t = 2\sec $ , we will use the concept of integration and derivations.

Complete step by step answer:
We have given that, velocity depends upon displacement as $v = 2x$ and we also know velocity can be expressed as derivative of displacement with respect to time so we have $v = \dfrac{{dx}}{{dt}}$
Now, put the value of $v = 2x$ we get,
$\dfrac{{dx}}{{dt}} = 2x$
Rearranging and solving above equation
$\dfrac{{dx}}{{2x}} = dt$
Integrating both sides we get,
$\int {\dfrac{{dx}}{{2x}}} = \int {dt} $
Let the limit of time starts with $t = 0$ and up to time $t$
So, limits of $x$ will be form $x = 2$ to some value of $x$
$\int\limits_2^x {\dfrac{{dx}}{x}} = 2\int\limits_0^t {dt} $
$\Rightarrow \ln (\dfrac{x}{2}) = 2t$
Integration of $\dfrac{1}{x}$ is $\ln x$
$x = 2{e^{2t}} \to (i)$
Now, we have a function of $x$ with $t$ ,let us find the value of this $x$ at time $t = 2\sec $
We get,
$x = 2{e^4}$ Which is simply the value of $x$ at time $t = 2\sec $ and, we also need to find the velocity at $t = 2\sec $
Now, as we have, the position-velocity function as $v = 2x$
We will put the value of $x = 2{e^4}$ in above given equation and we will get,
$v = 4{e^4}$ Which is nothing but the value of velocity at time $t = 2\sec $ .

Hence, the correct option is A.

Note:We must remember some basic concept of integration and their formulas like: The integration of $\int {dt = t} $ and integration of $\int {\dfrac{1}{x}dx = \ln x} $ and while putting limits, initial value of a variable is its lower limit while final value of a variable is its upper limit. $\int\limits_a^b {f'(x)dx} = f(b) - f(a)$.