Let's analyze the data and the statements.

The table gives the position of the car at different times:

We'll compute the average velocity over successive 2-second intervals:

1. From $t = 1 \, s$ to $t = 3 \, s$: $v_{\text{avg1}} = \frac{28 \, \text{m} - 10 \, \text{m}}{3 \, \text{s} - 1 \, \text{s}} = \frac{18}{2} = 9 \, \text{m/s}$
2. From $t = 3 \, s$ to $t = 5 \, s$: $v_{\text{avg2}} = \frac{44 \, \text{m} - 28 \, \text{m}}{5 \, \text{s} - 3 \, \text{s}} = \frac{16}{2} = 8 \, \text{m/s}$
3. From $t = 5 \, s$ to $t = 7 \, s$: $v_{\text{avg3}} = \frac{58 \, \text{m} - 44 \, \text{m}}{7 \, \text{s} - 5 \, \text{s}} = \frac{14}{2} = 7 \, \text{m/s}$
4. From $t = 7 \, s$ to $t = 9 \, s$: $v_{\text{avg4}} = \frac{70 \, \text{m} - 58 \, \text{m}}{9 \, \text{s} - 7 \, \text{s}} = \frac{12}{2} = 6 \, \text{m/s}$

The average velocity is decreasing over each interval, indicating the car is slowing down. Since the velocity is decreasing, the acceleration is negative. However, the car is still moving to the right (as the position is increasing), but the negative acceleration suggests deceleration to the right.

Now, evaluating the answer choices:

• A: Incorrect. The car is decelerating to the right, not accelerating to the left.
• B: Incorrect. Although the velocity is decreasing, it doesn't indicate acceleration to the left; it's deceleration to the right.
• C: Correct. The car is decelerating, but the velocity remains positive, which implies deceleration to the right (negative acceleration).
• D: Incorrect. This is a misleading explanation, as positive position alone doesn't determine acceleration direction.

So, the correct answer is C.

Would you like more details or have any further questions? Here are five related questions for deeper understanding:

1. How do you calculate average velocity in non-uniform motion?
2. What is the difference between average velocity and instantaneous velocity?
3. How does negative acceleration relate to deceleration?
4. Can an object have a positive velocity and negative acceleration at the same time?
5. How does the slope of a position-time graph represent velocity?

Tip: Deceleration doesn't always mean "slowing down in the reverse direction"; it refers to a reduction in speed, even if moving forward!