The speedometer of a car marks 55 km/h when the brakes are used. If the car completely stops in 3.9 seconds, determine the acceleration of the car.

To determine the acceleration of the car, we can use the following kinematic equation:

\[
v_f = v_i + at
\]

where:
- \( v_f \) is the final velocity (which is 0 since the car stops),
- \( v_i \) is the initial velocity (55 km/h),
- \( a \) is the acceleration,
- \( t \) is the time taken (3.9 seconds).

### Step 1: Convert the initial velocity to meters per second

The initial velocity is given as 55 km/h. We convert this to meters per second:

\[
v_i = 55 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 15.28 \, \text{m/s}
\]

### Step 2: Use the kinematic equation to solve for acceleration

The final velocity \( v_f \) is 0 m/s, the initial velocity \( v_i \) is 15.28 m/s, and the time \( t \) is 3.9 seconds. Substituting these values into the equation:

\[
0 = 15.28 \, \text{m/s} + a \times 3.9 \, \text{s}
\]

Rearrange the equation to solve for \( a \):

\[
a = \frac{-15.28 \, \text{m/s}}{3.9 \, \text{s}} = -3.92 \, \text{m/s}^2
\]

### Answer:
The acceleration of the car is \(-3.92 \, \text{m/s}^2\). The negative sign indicates that the car is decelerating.

Would you like to explore any particular aspect of this problem further? Here are some related questions that might interest you:

1. What is the relationship between acceleration and deceleration?
2. How do you interpret the negative sign in acceleration?
3. What is the difference between uniform acceleration and non-uniform acceleration?
4. How does time affect the deceleration of a car?
5. What are other kinematic equations used in motion analysis?

**Tip:** When dealing with deceleration, remember that acceleration will have a negative value, indicating a decrease in speed.

Discover how to determine the acceleration of a car using its initial speed and stopping time. This math problem involves converting units, applying the kinematic equation v_f = v_i + at, and understanding deceleration. Perfect for students in Grades 10-12.

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