To solve the problem, we need to analyze the motion of a particle moving along the x-axis with an initial velocity \( u > 0 \) and a constant negative acceleration \( a < 0 \). We are interested in finding the number of possible times the particle can be at a distance \( d \) from its starting point.
### Step-by-Step Solution:
1. **Understanding the Motion**:
The particle starts from the origin (0,0) with an initial velocity \( u \) and is decelerating due to negative acceleration \( a \). The motion can be described by the kinematic equation:
\[
x = ut + \frac{1}{2} a t^2
\]
where \( x \) is the position of the particle at time \( t \).
2. **Setting Up the Equation**:
We want to find the times \( t \) when the particle is at a distance \( d \) from the starting point, which gives us the equation:
\[
d = ut + \frac{1}{2} a t^2
\]
Rearranging this, we have:
\[
\frac{1}{2} a t^2 + ut - d = 0
\]
3. **Analyzing the Quadratic Equation**:
This is a quadratic equation in the form \( At^2 + Bt + C = 0 \), where:
- \( A = \frac{1}{2} a \)
- \( B = u \)
- \( C = -d \)
4. **Finding the Discriminant**:
The number of solutions (i.e., possible times \( t \)) depends on the discriminant \( D \) of the quadratic equation:
\[
D = B^2 - 4AC = u^2 - 4 \left(\frac{1}{2} a\right)(-d) = u^2 + 2ad
\]
5. **Determining the Number of Solutions**:
- If \( D > 0 \), there are **two distinct solutions** for \( t \).
- If \( D = 0 \), there is **one solution** for \( t \).
- If \( D < 0 \), there are **no real solutions** for \( t \).
6. **Conditions for \( D \)**:
Since \( a < 0 \), the term \( 2ad \) is negative. Therefore, the value of \( u^2 + 2ad \) can be analyzed:
- If \( u^2 > -2ad \), then \( D > 0 \) (two solutions).
- If \( u^2 = -2ad \), then \( D = 0 \) (one solution).
- If \( u^2 < -2ad \), then \( D < 0 \) (no solutions).
7. **Conclusion**:
The possible values for the number of times the particle can be at distance \( d \) from the starting point are:
- **1 time** (if it just touches the distance \( d \)),
- **2 times** (if it crosses the distance \( d \) while moving forward and then backward),
- **3 times** (if it moves past \( d \) and returns).
Thus, the final answer is that the number of possible values of time for which the particle is at distance \( d \) from its starting point may be **1, 2, or 3**.
