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Constant velocity is a central concept in kinematics, often referred to as uniform motion. It occurs when an object travels the same distance every second without changing its speed. This means both the speed and direction of the object remain consistent over time.

A practical application of constant velocity can be observed in the given problem, where the cockroach is moving at a constant speed of 1.50 m/s. Since its velocity is constant, there is no acceleration acting on the cockroach. This simplifies calculations because the primary relationship we need to consider is the equation for velocity:

• Displacement (\( s \)) is calculated by multiplying velocity (\( v \)) by time (\( t \)).
• Therefore, \( s = vt \).

Understanding this allows us to calculate how long the cockroach will travel to cover a particular displacement, without needing to consider changes in speed.

Displacement is a vector quantity that refers to the change in position of an object. It is not just about how far the object has traveled, but rather the overall change from the starting point to the end point in a straight line.

In the exercise, the concept of displacement is crucial. The cockroach needs to travel 1.20 meters before reaching safety. Thus, your task is to cover a total displacement of 2.10 meters because you start 0.90 meters behind the cockroach. The total displacement incorporates two parts:

• Initial gap: 0.90 meters, which is the distance between you and the cockroach at the start.
• Cockroach's target displacement: 1.20 meters.

In solving problems like this, understanding how to calculate total displacement guides you toward finding how quickly and efficiently the distance can be covered by each moving entity.

Acceleration is the rate of change of velocity of an object. It describes how quickly an object speeds up, slows down, or changes direction over time. In kinematic problems, finding acceleration involves understanding the balance between initial speed, time, and distance traveled.

For the problem at hand, you start with an initial speed of 0.80 m/s and need to accelerate to catch the cockroach. The formula used to calculate this needed acceleration is derived from the equation:

• \( s = ut + \frac{1}{2}at^2 \)
• where \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time.

This equation takes into account the initial speed and the required acceleration to cover the necessary displacement of 2.10 meters within the same time frame, 0.80 seconds, in which the cockroach travels its own path.

Kinematic equations provide a way to calculate various parameters of motion, such as displacement, velocity, acceleration, and time, when the acceleration is constant. These equations are essential tools in solving problems in kinematics.

In the given problem, the use of kinematic equations allows for determining the acceleration required to catch the cockroach. The key kinematic equation used is:

• \( s = ut + \frac{1}{2}at^2 \)
• This equation helps solve for \( a \) by setting \( s \) (total displacement), \( u \) (initial speed), and \( t \) (time calculated from the cockroach's motion).

Other kinematic equations, like \( v = u + at \) or \( v^2 = u^2 + 2as \), provide further ways to handle different kinematic scenarios. In this particular problem, selecting the appropriate equation simplifies finding the exact acceleration needed without trial and error.