91Ó°ÊÓ

Constant acceleration is a key concept in physics. It means that the velocity of an object changes at a steady rate over time. For the given problem, the car has a constant acceleration of 4.0 m/s².
When an object starts from rest (initial velocity = 0) and moves with constant acceleration, we can calculate its position using the equation:

\(s = ut + \frac{1}{2}at^2\)

This formula combines the effects of initial velocity and the acceleration component over time. Due to the constant acceleration, the car's velocity increases linearly.
Keep in mind that units should always be consistent. For example, acceleration should be in meters per second squared (m/s²), time in seconds (s), and distance in meters (m).
In our problem, this equation helps in finding how far the car travels in a given time, such as 3 seconds.

Equations of motion are essential for solving problems involving moving objects. They describe the relationship between an object's velocity, position, and acceleration over time. Let's focus on the three primary equations:

• v = u + at
• s = ut + \(\frac{1}{2}at^2\)
• v² = u² + 2as

In the given exercise, we use the second equation to find the car's position after 3 seconds. Here,

• u is the initial velocity,
• a is constant acceleration,
• t is time,
• s is the position.

For the truck, which travels at a constant speed, the position is calculated using the simple formula:
\(s = vt\)
Here, v is the constant speed, and t is the time. Understanding these formulas allows you to find how far the car and truck have traveled from the traffic light at any given time.

Determining the center of mass is crucial when dealing with a system of multiple objects. The center of mass gives a single point where the entire mass of the system is considered to be concentrated.
For our problem, the center of mass of the car and truck can be calculated with:

\(x_{cm} = \frac{m1 \cdot x1 + m2 \cdot x2}{m1 + m2}\)

This formula represents a weighted average of the positions of the car and truck:

• m1 and m2 are the masses of the car and truck respectively,
• x1 and x2 are their positions.

By plugging in the appropriate values, we can find the center of mass at any point in time.
For instance, in 3 seconds, we get:

\(x_{cm} = \frac{1000 \times 18 + 2000 \times 24}{1000 + 2000} = 22 \, m\)

This means the center of mass is 22 meters from the traffic light.

Velocity is a measure of how fast an object is moving in a specific direction. It has both magnitude and direction, making it a vector quantity.
In the given scenario, the car starts from rest and speeds up due to constant acceleration. Its velocity after a certain time can be found using:

\(v = u + at\)

Since the car starts from rest, u = 0, so the formula simplifies to:

\(v = at\)

For the truck, the velocity remains constant at 8.0 m/s.
Understanding these differences in motion helps determine how fast both vehicles travel and their effect on the system's overall speed. At 3 seconds, the car's velocity becomes:

\(v = 4.0 \, m/s^2 \times 3 \, s = 12 \, m/s\)

Hence, the truck's speed is unaffected, but the car speeds up steadily.

Position calculation is fundamental in physics to determine where an object is at any given time. With motion involving acceleration, the position of an object can be found using certain equations.
For example, in a system where one object accelerates and another moves at constant speed, both formulas from the earlier sections apply:

• For acceleration: \(s = ut + \frac{1}{2}at^2\)
• For constant speed: \(s = vt\)

In our problem, we calculated the position of:

• The car using \(s = 0 + \frac{1}{2} \times 4 \times (3)^2 = 18 \, m\)
• The truck using \(s = 8 \times 3 = 24 \, m\)

Understanding and applying these equations help in accurately determining objects' positions, which is essential in analyzing real-world problems involving multiple moving entities.