Answer

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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 $\dfrac{v_0}{\alpha}$.

**Note:**

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.

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