**Problem Analysis:**

Let's assume that the first car starts at time t = 0 and the second car starts at time t = 1 minute = 60 seconds. Both cars have an acceleration of 0.2 m/s^2.

We need to find the time when the distance between the two cars becomes equal to three times its initial value (when the second car just starts).

**Solution:**

To solve this problem, we can use the equations of motion. The equation relating distance, time, and acceleration is:

\[s = ut + \frac{1}{2}at^2\]

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Let's consider the first car:

- Initial velocity (u1) = 0 m/s (as the car starts from rest)
- Acceleration (a1) = 0.2 m/s^2
- Time (t) = t

The distance traveled by the first car (s1) after time t is:

\[s1 = u1t + \frac{1}{2}a1t^2\]
\[s1 = \frac{1}{2}a1t^2\] (since u1 = 0)

Now, let's consider the second car:

- Initial velocity (u2) = 0 m/s (as the car starts from rest)
- Acceleration (a2) = 0.2 m/s^2
- Time (t) = t - 60 seconds (since the second car starts 1 minute = 60 seconds later)

The distance traveled by the second car (s2) after time t is:

\[s2 = u2(t - 60) + \frac{1}{2}a2(t - 60)^2\]
\[s2 = \frac{1}{2}a2(t - 60)^2\] (since u2 = 0)

Now, let's find the time when the distance between the two cars becomes three times its initial value:

\[s2 - s1 = 3s1\]
\[\frac{1}{2}a2(t - 60)^2 - \frac{1}{2}a1t^2 = 3(\frac{1}{2}a1t^2)\] (substituting s1 and s2)

Simplifying the equation:

\[a2(t - 60)^2 - a1t^2 = 3a1t^2\]
\[a2(t^2 - 120t + 3600) - 4a1t^2 = 0\]
\[(a2 - 4a1)t^2 - a2(120t - 3600) = 0\]

Since this is a quadratic equation in t, we can solve it by factoring or using the quadratic formula. However, this quadratic equation does not have any real solutions. Therefore, there is no time when the distance between the two cars becomes three times its initial value.

Hence, the answer to the problem is that there is no such time.