To solve this problem, we can use the following kinematic equation:

$v^2 = u^2 + 2as$

Where:

• $v$ is the final velocity (what we are trying to find),
• $u$ is the initial velocity ($u = 20 \, \text{m/s}$),
• $a$ is the acceleration ($a = 5 \, \text{m/s}^2$),
• $s$ is the displacement ($s = 400 \, \text{m}$).

Let's plug in the values and solve for $v$:

$v^2 = (20)^2 + 2(5)(400)$ $v^2 = 400 + 4000$ $v^2 = 4400$ $v = \sqrt{4400} \approx 66.33 \, \text{m/s}$

So, the car will be moving at approximately 66.33 m/s after traveling 400 meters.

Would you like more details on any step, or do you have any other questions?

Here are some related questions you might explore:

1. How long will it take for the car to travel the 400 meters?
2. What would the velocity be if the acceleration was 3 m/s² instead?
3. How far will the car travel in 10 seconds under this acceleration?
4. If the car decelerates at 5 m/s², how long will it take to stop?
5. What is the car’s average velocity over the 400-meter distance?

Tip: When solving kinematic problems, always ensure you correctly identify the known variables and choose the appropriate equation accordingly.