A jetliner touches down at \(220 \mathrm{km} / \mathrm{h}\) and comes to a halt \(29 \mathrm{s}\)

Understanding uniformly accelerated motion is crucial when analyzing objects that are speeding up or slowing down at a constant rate. When an object, like the jetliner in the exercise, comes to a halt, it decelerates uniformly because the changes in its velocity occur at a constant rate over time. The motion can be described using kinematic equations which apply to any moving object experiencing uniform acceleration (or deceleration).

It's important to note that uniformly accelerated motion doesn't mean the object's speed is uniform! It means the acceleration is constant, not the velocity. This constant acceleration is what allows us to use specific kinematic equations to predict future motion, such as how much distance it takes for the jetliner to stop.

Converting units is a fundamental skill in physics which ensures that we work with standardized measures and can meaningfully compare results. In our jetliner example, converting from kilometers per hour to meters per second makes it possible to use the SI unit system, which is the standard for scientific calculations. To convert, we multiply the velocity by the factor \( \frac{5}{18} \), because one hour has 3600 seconds and one kilometer has 1000 meters. This conversion is essential to accurately calculate acceleration and distance in the later steps.

Always check if your units match the formulas you’re going to use, as mixing units can lead to incorrect results.

Velocity and acceleration are two fundamental concepts in kinematics. Velocity describes the speed and direction of an object, while acceleration measures how quickly the velocity changes. In our exercise, the initial velocity (\(u\)) is the jetliner's speed when it touches down, and the final velocity (\(v\)) is zero because the jetliner comes to a halt.

Deceleration occurs when acceleration is negative, meaning the object is slowing down. By rearranging the kinematic equation \(v = u + at\), we're able to calculate the acceleration, which is essential when determining the distance required for the jetliner to stop.

Distance calculation in physics often involves determining how far an object travels during a certain period. Our exercise uses the kinematic equation \(s = ut + \frac{1}{2} a t^2\), where \(s\) represents the distance covered. It's a combination of the distance the jetliner would cover with no acceleration (\(ut\)) and the additional distance due to constant acceleration (\(\frac{1}{2} a t^2\)).

When calculating this equation, all values must be in consistent units, which is why we converted our initial velocity. This equation helps us predict how much runway the jetliner needs to come to a stop after touchdown, a critical aspect of airport design and safety planning.