To solve the problem of finding the distance traveled by the car while it was accelerating, we can use one of the kinematic equations that relates initial velocity, final velocity, acceleration, and distance. The appropriate equation for this scenario is:

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

Where:

• $v$ is the final velocity (22 m/s),
• $u$ is the initial velocity (14 m/s),
• $a$ is the acceleration (0.85 m/s²),
• $s$ is the distance traveled.

We need to solve for $s$:

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

Now, substitute the values:

$s = \frac{(22)^2 - (14)^2}{2 \times 0.85}$

$s = \frac{484 - 196}{1.7}$

$s = \frac{288}{1.7} \approx 169.41 \text{ meters}$

So, the car traveled approximately 169.41 meters while it was accelerating.

Would you like more details or have any questions?

Here are five related questions:

1. How would the distance change if the acceleration were increased to 1.5 m/s²?
2. What would be the time taken for the car to reach the final velocity?
3. If the car continued at the final velocity for 10 more seconds, what would be the additional distance traveled?
4. How would the distance traveled change if the initial velocity was 10 m/s instead of 14 m/s?
5. What would be the car's velocity after traveling 100 meters with the same acceleration?

Tip: When dealing with constant acceleration, kinematic equations are powerful tools for solving problems involving distance, velocity, and time.