Solution:

To solve this, we'll use the equation of motion:

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

Where:
- \( v = 0 \) (final velocity, since the car comes to rest)
- \( u = 144 \, \text{km/h} = 40 \, \text{m/s} \) (initial velocity)
- \( s = 200 \, \text{m} \) (distance)
- \( a \) is the acceleration (which we'll solve for)

Rearranging the equation:

\[
0 = (40)^2 + 2 \cdot a \cdot 200
\]
\[
0 = 1600 + 400a
\]
\[
a = - \frac{1600}{400} = -4 \, \text{m/s}^2
\]

Now, to find the time (\( t \)):

\[
v = u + at
\]
\[
0 = 40 + (-4)t
\]
\[
t = \frac{40}{4} = 10 \, \text{seconds}
\]

Thus, it takes 10 seconds for the car to stop.