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The problem describes a car decelerating with an initial velocity of \( v_0 = 25 \, \text{m/s} \), a deceleration (negative acceleration) of \( a = -1.8 \, \text{m/s}^2 \), and a distance traveled of \( d = 36 \, \text{m} \). The goal is to find the time \( t \) taken to cover this distance.

We can use the following kinematic equation to solve for \( t \):

\[
d = v_0 t + \frac{1}{2} a t^2
\]

### Known values:
- \( v_0 = 25 \, \text{m/s} \)
- \( a = -1.8 \, \text{m/s}^2 \)
- \( d = 36 \, \text{m} \)

We now substitute these values into the equation:

\[
36 = 25t + \frac{1}{2}(-1.8)t^2
\]

This simplifies to:

\[
36 = 25t - 0.9t^2
\]

Rearranging into a standard quadratic form:

\[
0.9t^2 - 25t + 36 = 0
\]

I will now solve this quadratic equation to find the time \( t \).The two solutions to the quadratic equation are:

- \( t_1 = 26.2542 \, \text{seconds} \)
- \( t_2 = 1.5236 \, \text{seconds} \)

Since time cannot be negative or too large for this scenario, the correct solution is \( t = 1.5236 \, \text{seconds} \).

To match the requirement of two decimals without rounding, the final answer is:

**1.52**

Would you like further details or clarifications? Here are five related questions for further understanding:

1. What is the role of the negative acceleration in this problem?
2. How would the equation change if the car were speeding up instead of slowing down?
3. What other kinematic equations could be used if the final velocity were provided?
4. How does the discriminant of the quadratic formula affect the number of real solutions?
5. Why is it important to discard the larger time in this context?

**Tip:** Always double-check whether the given values represent acceleration or deceleration to correctly assign the sign of the acceleration.

A car travelling at 25 m/s starts slowing down at a rate of 1.8 m/s^2. In how much time will it travel 36 m? Express your answer in seconds. Only give the numerical answer. No units. Use at least 4 decimals in the procedure. Provide only 2 decimals in the final result, without rounding.

Learn how to calculate the time it takes for a car to travel 36 meters while decelerating from 25 m/s at a rate of 1.8 m/s². This problem involves kinematic equations and solving a quadratic equation. Suitable for students in Grades 10-12.

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