A particle moves with initial velocity $v_0$ and retardation $\alpha v$, where $v$ is its initial velocity at any time t and $\alpha$ is a constant.
A. The particle will cover a total distance $\dfrac{v_0}{\alpha}$.
B. The particle will come to rest after a time $\dfrac{1}{\alpha}$.
C. The particle will continue to move for a very long time.
D. The velocity of the particle will become $\dfrac{v_0}{2}$ after a time $\dfrac{1}{\alpha}$.

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Hint: It is given that the acceleration (retardation) is proportional to velocity. Therefore, a separable form of a differential equation is already apparent if we write this condition down, as the equations of motion are nothing but differential equations.

Formula Used:
The acceleration of a body is related to its position as:
$a= \dfrac{d^2x}{dt^2}$

Complete step by step answer:
Retardation is caused when a body is undergoing a decrease in velocity with time. Retardation is just opposite to acceleration. We are given that our particle undergoes retardation $\alpha v$. Mathematically we may write:
$a= \dfrac{d^2x}{dt^2} = - \alpha v$,
Since the RHS has velocity, we try to make a velocity term on the LHS too.
$\dfrac{dv}{dt} = - \alpha v $
On the RHS, we multiply and divide by dx, we get:
$\dfrac{dv . dx}{dt . dx} = - \alpha v $
which just becomes
$v. \dfrac{dv }{ dx} = - \alpha v $
It becomes visible now that the expression only has x and v variables remaining.
Canceling the v on both sides, we get:
$ \dfrac{dv }{ dx} = - \alpha $
$dv = - \alpha dx$
we have the parameters:
At t=0, velocity is $v_0$ and we assume that x=0 and after a time t, the velocity will become zero and it would have covered some distance x.
Taking integral on both sides and keeping the limits:
$\int_{0}^{v_0} dv = \int_{x}^{0} - \alpha dx $
We get:
$(0-v_0) = - \alpha (x-0)$
Which gives us the result that the particle after a time t, covers a distance
$x= \dfrac{v_0}{\alpha}$.
Therefore, the correct answer appears to be (A). The particle will cover a total distance $\dfrac{v_0}{\alpha}$.

The key in the question is the differential equation. Analyze it carefully so that the difficulty decreases later. Here, we eliminated the dt form and rather constructed a differential with respect to position so that it becomes a linear differential equation of first order.