To solve this problem, we need to calculate the acceleration of the car.

Given:

1. Initial velocity ($v_0$) = 10 m/s
2. Final velocity ($v$) = 0 m/s (the car stops)
3. Distance ($s$) = 100 m

Formula for acceleration:

We use the equation of motion: $v^2 = v_0^2 + 2as$ Where:

• $v$ is the final velocity,
• $v_0$ is the initial velocity,
• $a$ is the acceleration,
• $s$ is the displacement.

Rearrange to solve for $a$:

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

Substituting the values:

$a = \frac{0^2 - (10)^2}{2 \times 100} = \frac{-100}{200} = -0.5 \, \text{m/s}^2$

Answer:

The acceleration of the car is: $\mathbf{-0.5 \, \text{m/s}^2}$

Thus, the correct option is: D. -0.5 m/s²

Would you like further explanation or details? Here are 5 related questions to consider:

1. What does negative acceleration signify in this context?
2. How would the answer change if the initial velocity were different?
3. What is the relationship between acceleration, velocity, and displacement?
4. Can you calculate the time it takes for the car to stop using this acceleration?
5. How does the equation $v^2 = v_0^2 + 2as$ relate to real-world motion?

Tip: Remember, negative acceleration (deceleration) occurs when an object slows down!