To solve this problem, we can use one of the kinematic equations for motion under constant acceleration:

$s = v_0 t + \frac{1}{2} a t^2$

where:

• $s$ is the displacement (what we're solving for),
• $v_0$ is the initial velocity (0 m/s, since the car starts from rest),
• $a$ is the acceleration (5 m/s$^2$),
• $t$ is the time (3.6 seconds).

Since $v_0 = 0$, the equation simplifies to:

$s = \frac{1}{2} a t^2$

Now, substitute the given values into the equation:

$s = \frac{1}{2} \times 5 \, \text{m/s}^2 \times (3.6 \, \text{s})^2$

$s = \frac{1}{2} \times 5 \times 12.96 = 2.5 \times 12.96 = 32.4 \, \text{m}$

So, the car's displacement over the period of 3.6 seconds is 32.4 meters.

Would you like more details on the steps involved, or do you have any questions?

Here are 5 related questions:

1. What would be the final velocity of the car at the end of the 3.6 seconds?
2. How much time would it take for the car to reach a displacement of 100 meters under the same acceleration?
3. If the car maintained the final velocity from the acceleration phase, how far would it travel in the next 5 seconds?
4. How would the displacement change if the acceleration were halved?
5. What is the car's average velocity during the acceleration phase?

Tip: In problems involving acceleration from rest, the displacement is directly proportional to the square of the time.