Ω and C = 10μF.

The circuit diagram is as follows:

```
+---------R---------+
| |
V_in C
| |
+--------------------+
|
V_out
```

We can analyze the circuit using the following steps:

1. Find the expression for the output voltage V_out(t) in terms of the input voltage V_in(t).
2. Calculate the time constant τ = RC.
3. Find the derivative of the input voltage dv/dt.
4. Find the maximum output voltage V_out(max) that occurs when dv/dt is maximum.
5. Find the time t_max at which V_out(max) occurs.

Step 1:

The output voltage V_out(t) is given by the formula:

V_out(t) = -RC(dV_in/dt)

Substituting the given values of R and C, we get:

V_out(t) = -50(dV_in/dt)

Step 2:

The time constant τ = RC = 5kΩ x 10μF = 50ms.

Step 3:

The derivative of the input voltage V_in(t) is given by:

dV_in/dt = 1V/s

Step 4:

The maximum output voltage V_out(max) occurs when the derivative dv/dt is maximum. Since dv/dt is constant in this case, V_out(max) is simply:

V_out(max) = -50(dV_in/dt) = -50(1V/s) = -50V/s

Step 5:

The time t_max at which V_out(max) occurs can be found by setting the derivative of V_out(t) to zero and solving for t:

dV_out/dt = -50(d^2V_in/dt^2) = 0

d^2V_in/dt^2 = 0

Since dV_in/dt is constant, its second derivative is zero. Therefore, t_max can be any value.

Therefore, the output voltage V_out(t) is:

V_out(t) = -50(dV_in/dt) = -50(1V/s) = -50V/s

And it remains constant for all values of t.