To find the distance the car travels during the acceleration, we can use the following kinematic equation:

$d = v_i t + \frac{1}{2}at^2$

Since the car starts from rest, the initial velocity $v_i$ is 0, simplifying the equation to:

$d = \frac{1}{2}at^2$

First, we need to find the acceleration $a$, using the formula:

$a = \frac{v_f - v_i}{t}$

Here:

• $v_i = 0 \, \text{m/s}$ (initial velocity, since the car starts from rest)
• $v_f = 6.58 \, \text{m/s}$ (final velocity)
• $t = 6.5 \, \text{seconds}$ (time)

Substituting the values:

$a = \frac{6.58 \, \text{m/s} - 0 \, \text{m/s}}{6.5 \, \text{seconds}} = \frac{6.58 \, \text{m/s}}{6.5 \, \text{seconds}} \approx 1.012 \, \text{m/s}^2$

Now, substituting $a$ and $t$ into the displacement equation:

$d = \frac{1}{2} \times 1.012 \, \text{m/s}^2 \times (6.5 \, \text{seconds})^2$

$d = 0.506 \, \text{m/s}^2 \times 42.25 \, \text{seconds}^2$

$d \approx 21.39 \, \text{meters}$

Rounding to the nearest whole number, the car travels 21 meters during this time.

Would you like further details or have any questions?

Here are some related questions you might consider:

1. What would the distance be if the car accelerated for 10 seconds?
2. How much time would it take to reach 10 m/s with the same acceleration?
3. What is the car’s average velocity during the acceleration?
4. How would the distance change if the final speed was 10 m/s?
5. Can you determine the acceleration if the car reached 20 meters in 6.5 seconds?

Tip: When rounding off your answer, ensure that all intermediate steps maintain sufficient decimal places to avoid errors in the final result.