A

To solve the problem, we start by recognizing that the equation governing the motion under retardation (deceleration) can be modeled by the differential equation:
$$ \frac{dv}{dt} = -kv $$
where $v$ is the velocity, $k$ is a constant, and $t$ is time.
Now, we can separate the variables and integrate:
$$ \int \frac{1}{v} dv = -k \int dt $$
This yields:
$$ \ln |v| = -kt + C $$
Solving for $v$, we exponentiate both sides:
$$ v = e^{-kt + C} = e^C e^{-kt} = u e^{-kt} $$
where $u$ is the initial velocity when $t=0$.
The displacement $s$ can be found by integrating the velocity:
$$ s = \int v \, dt = \int u e^{-kt} dt = -\frac{u}{k} e^{-kt} + C' $$
Evaluating this gives:
$$ s = -\frac{u}{k} (1 - e^{-kt}) $$
Now, substituting $t$ into this equation, we get the distance covered in $t$ seconds.
Thus, confirming our calculations and the final expression aligns with Option A.
Therefore, the correct choice is A.