when trying to solve it, o got -10.7 due to me inputting acceleration as a negative because it is slowing. Is this not correct?

Your approach of inputting acceleration as negative is indeed correct when considering the deceleration of the truck. The key is in understanding how the equation of motion works in this context.

Let's revisit the equation of motion that relates velocity, acceleration, and distance:
\[
v^2 = u^2 + 2as
\]

Where:
- \( v \) is the final velocity (0 m/s in this case),
- \( u \) is the initial velocity (13.89 m/s),
- \( a \) is the acceleration (or deceleration, which is negative),
- \( s \) is the distance traveled.

Given:
- \( u = 13.89 \, \text{m/s} \)
- \( v = 0 \, \text{m/s} \)
- \( a = -9 \, \text{m/s}^2 \)

When solving for distance \( s \), the equation becomes:
\[
0 = (13.89)^2 + 2(-9)s
\]
\[
0 = 192.93 - 18s
\]
\[
18s = 192.93
\]
\[
s = \frac{192.93}{18} \approx 10.72 \, \text{m}
\]

**Important Note:**
- The distance \( s \) should always be a positive value because it represents a physical distance.
- The negative sign in the acceleration term already accounts for the fact that the truck is slowing down (decelerating).

If you were to incorrectly substitute acceleration as negative in a way that introduces an additional negative sign on the right-hand side, you might incorrectly calculate a negative distance:
\[
0 = 192.93 + 18s  \quad \text{(incorrect)}
\]
\[
s = -\frac{192.93}{18} \approx -10.72 \, \text{m} \quad \text{(incorrect)}
\]
This would yield an unphysical negative distance.

### Correct Approach Recap:
1. **Use the correct equation**: \( v^2 = u^2 + 2as \).
2. **Keep the acceleration negative** for deceleration.
3. **Solve for distance \( s \)**, ensuring that the result is a positive value since distance cannot be negative.

Your method was on the right track by considering the negative acceleration, but it's important to apply it correctly within the equation to ensure the correct result. The final distance \( s \) should indeed be positive.

Would you like any further clarification?

Here are 5 related questions:
1. How does the sign of acceleration affect the interpretation of the motion?
2. What are other scenarios where a negative acceleration (deceleration) occurs?
3. How would you apply this method to calculate stopping distances for different vehicles?
4. How does initial speed impact stopping distance when deceleration remains constant?
5. What role does reaction time play in determining total stopping distance?

**Tip:** Always double-check the signs in your equations, especially when dealing with deceleration, to ensure physically meaningful results.

Learn how to correctly calculate distance when dealing with deceleration (negative acceleration) in physics. This problem explores the equation of motion v^2 = u^2 + 2as, where u is initial velocity, v is final velocity, a is acceleration (negative for deceleration), and s is distance. Get step-by-step guidance and avoid common mistakes in physics calculations.

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