If d denotes the distance covered by a car in time t and `vecs` denotes the displacement by the car during the same time, then :

Options

A. `dle|vecS|`
B. `d=|vecS|` only
C. `dge|vecS|`
D. `dle|vecS|`

Explanation

To solve the problem regarding the relationship between distance (d) and displacement (s) for a car moving in time t, we can follow these steps:

### Step-by-Step Solution:

1. **Define Distance and Displacement**:
- **Distance (d)**: The total length of the path traveled by the car, regardless of the direction.
- **Displacement (s)**: The shortest straight-line distance from the initial position to the final position of the car, along with the direction.

2. **Understanding the Relationship**:
- Displacement is always less than or equal to distance because it represents the shortest path between two points.
- In mathematical terms, we can express this as:
\[ |s| \leq d \]
where \( |s| \) is the magnitude of displacement.

3. **Illustration with an Example**:
- Consider a car moving from point A to point C via points B. If the car takes the path A → B → C, the distance covered (d) is the total length of the path A to B plus B to C.
- However, if the car moved directly from A to C, the displacement (s) would be the straight-line distance from A to C.

4. **Conclusion**:
- Therefore, we conclude that the magnitude of displacement is always less than or equal to the distance traveled by the car. This can be summarized as:
\[ |s| \leq d \]

### Final Statement:
Thus, the relationship between distance and displacement can be expressed as:
- The magnitude of displacement (|s|) is less than or equal to the distance (d).