**Magnitude of Acceleration:**

To find the magnitude of the bullet's acceleration, we can use the kinematic equation:

\[v^2 = u^2 + 2as\]

Where:
- \(v\) is the final velocity (0 m/s, since the bullet stops)
- \(u\) is the initial velocity (150 m/s)
- \(a\) is the acceleration (unknown)
- \(s\) is the distance traveled before stopping (3.5 cm or 0.035 m)

Rearranging the equation to solve for acceleration, we have:

\[a = \frac{{v^2 - u^2}}{{2s}}\]

Substituting the given values, we have:

\[a = \frac{{0^2 - 150^2}}{{2 \times 0.035}}\]

Simplifying the equation:

\[a = \frac{{-150^2}}{{0.07}}\]

Calculating the result:

\[a \approx -107142.857 \, \text{m/s}^2\]

Since the bullet is decelerating (slowing down), the acceleration has a negative sign.

**Time Taken to Stop:**

To find the time taken for the bullet to stop, we can use the equation:

\[v = u + at\]

Where:
- \(v\) is the final velocity (0 m/s)
- \(u\) is the initial velocity (150 m/s)
- \(a\) is the acceleration (-107142.857 m/s^2)
- \(t\) is the time taken (unknown)

Rearranging the equation to solve for time, we have:

\[t = \frac{{v - u}}{{a}}\]

Substituting the given values, we have:

\[t = \frac{{0 - 150}}{{-107142.857}}\]

Simplifying the equation:

\[t = \frac{{-150}}{{-107142.857}}\]

Calculating the result:

\[t \approx 0.001401 \, \text{s}\]

Therefore, it takes approximately 0.001401 seconds for the bullet to stop.

In summary, the magnitude of the bullet's acceleration is approximately 107142.857 m/s^2 (negative sign indicates deceleration), and it takes approximately 0.001401 seconds for the bullet to stop moving.