We can solve this using one of the kinematic equations of motion. The equation that relates initial velocity ($v_0$), final velocity ($v$), acceleration ($a$), and displacement ($d$) is:

$v^2 = v_0^2 + 2ad$

Where:

• $v_0 = 10 \, \text{m/s}$ (initial velocity),
• $v = 30 \, \text{m/s}$ (final velocity),
• $a = 22 \, \text{m/s}^2$ (acceleration),
• $d$ is the distance traveled (which we are trying to find).

Step 1: Rearrange the formula to solve for distance $d$:

$d = \frac{v^2 - v_0^2}{2a}$

Step 2: Plug in the known values:

$d = \frac{(30)^2 - (10)^2}{2 \times 22}$

Now let's compute that:

$d = \frac{900 - 100}{44} = \frac{800}{44} \approx 18.18 \, \text{meters}$

Final answer:

The car travels approximately 18.18 meters.

Do you need any further details or explanations?

Here are 5 related questions to consider:

1. How would the distance change if the acceleration was halved?
2. What time does the car take to reach its final velocity?
3. What is the average velocity of the car during this acceleration?
4. How much distance would the car cover if the initial velocity were zero?
5. What would happen to the distance if the final velocity were doubled?

Tip: Always keep the units consistent when solving physics problems to avoid errors in calculation!