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In physics, the notion of relative speed comes into play when determining how fast one object is moving in comparison to another. It is a critical concept for predicting interactions between moving objects, such as the possibility of collision in traffic scenarios.

When assesssing the relative speed between two objects, you subtract the speed of one from the speed of the other. For instance, in the example of Speedy Sue and the slow-moving van, Sue's speed is given as 30 m/s, while the van's speed is 5 m/s. Thus, the relative speed is calculated as:
$$30.0 \, \text{m/s} - 5.0 \, \text{m/s} = 25.0 \, \text{m/s}$$
This value represents how quickly Sue is approaching the van and is the first step in determining whether a collision could occur.

Deceleration refers to any decrease in speed or the negative acceleration of an object. In the context of our exercise, Sue's deceleration is caused by applying brakes on a wet road. It's essential to understand that deceleration reduces an object's speed at a certain rate over time, often measured in meters per second squared (m/s²).

The deceleration is given as $$-2.00 \, \text{m/s}^{2}$$This means that Sue's car slows down by 2 m/s every second. Deceleration plays a pivotal role in calculating the time taken for an object to come to a complete stop and the distance it covers during that time.

The distance-time calculation is a fundamental analysis in kinematics, the branch of physics that deals with motion. It tells us how far an object travels over a specific time period when moving at a constant speed.

In our scenario, once we know the time it would take for Sue to stop her car (15.0 s), we calculate the distance the van, which is moving at a constant velocity, will cover in this time using the simple formula:
$$\text{distance} = \text{speed} \times \text{time}$$
The van travels 75m in the 15s Sue is coming to a stop. This distance-time consideration is crucial in determining whether Sue and the van will have sufficient room to avoid a collision.

Predicting a collision involves analyzing the positions, speeds, and accelerations (or decelerations) of the moving objects. In kinematics problems, such predictions are pivotal in determining whether two objects will occupy the same position at the same time, leading to a collision. By considering the relative speed and the distance-time calculation, we predict the outcome of Sue's braking scenario.

Since the van will move only 75 meters while Sue comes to a stop, and considering they were 155 meters apart to begin with, they will not collide. The two will be closest at a distance of:$$155 \, \text{m} - 75 \, \text{m} = 80 \, \text{m}$$Hence, Sue's car will stop with 80 meters to spare before reaching the van. This distance of closest approach finalizes the prediction: no collision would occur.