91Ó°ÊÓ

Kinematics equations serve as powerful tools for analyzing motion with constant acceleration. These are a set of four equations used to describe the positions, velocities, and accelerations of objects:

• \( v = v_0 + a \cdot t \)
• \( s = s_0 + v_0 \cdot t + \frac{1}{2} a \cdot t^2 \)
• \( v^2 = v_0^2 + 2a(s - s_0) \)
• \( s = s_0 + \frac{1}{2}(v + v_0) \cdot t \)

These equations are essential because they allow us to calculate variables such as final velocity and displacement, given the initial conditions and constant acceleration. When a particle moves, like in our exercise, we adopt these formulas:

• For the x-direction: initial velocity \(v_{0x} = 0\), acceleration \(a_x = 2 \mathrm{~m/s^2}\)
• For the y-direction, beginning after 2 seconds: initial velocity \(v_{0y} = 0\), acceleration \(a_y = -4 \mathrm{~m/s^2}\)

Two-dimensional motion refers to the movement of an object in a plane, involving coordinates on both the x-axis and y-axis. Unlike one-dimensional motion that follows a straight path, two-dimensional motion examines how objects navigate through both axes. This complexity requires analyzing each component separately.
When a particle moves along both axes, like in our exercise, the analysis begins with separating the motion into x and y directions. For each direction, we use specific kinematics equations based on initial conditions and any applied accelerations. By analyzing these components independently, and then combining the results, we obtain a comprehensive understanding of the particle’s journey.
In the exercise example, the particle first accelerates in the positive x-direction for 4 seconds and then an acceleration acts in the negative y-direction starting after 2 seconds. This results in a combined motion path, which we decipher by analyzing individual parts and synthesizing the findings.

Velocity calculation in a two-dimensional framework involves assessing motion along two separate axes. For each axis, velocity is computed using specific initial conditions and accelerations. At time \(t=0\), the initial velocities for both axes are set to zero in our problem.
Surrounding the topic of velocity:

• For the x-axis, the velocity at 4 seconds: \(v_x = v_{0x} + a_x \times t = 0 + 2 \times 4 = 8 \mathrm{~m/s}\)
• For the y-axis, the velocity after 2 seconds of acting acceleration: \(v_y = v_{0y} + a_y \times (t-2) = 0 - 4 \times 2 = -8 \mathrm{~m/s}\)
• Overall velocity combines these components, calculated using Pythagoras’ theorem: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{8^2 + (-8)^2} = 8\sqrt{2} \mathrm{~m/s}\)

The result represents the particle's velocity in both x and y directions, giving a vector direction as well as magnitude, illustrating not just speed but the path and changes in motion.

Coordinate determination is key to understanding where an object is located at a given time. In two-dimensional motion, this involves computing the position coordinates on both the x and y axes. Given constant acceleration and initial conditions, we use kinematics equations to find precise positions.
For the x and y coordinates in our exercise:

• The x-position, given initial position and x-axis motion: \(s_x = x_0 + v_{0x} \times t + \frac{1}{2} a_x \times t^2 = 2 + 0 + \frac{1}{2} \cdot 2 \cdot 4^2 = 18 \mathrm{~m}\)
• The y-position, accounting for acceleration start time after 2 seconds: \(s_y = y_0 + v_{0y} \times (t-2) + \frac{1}{2} a_y \times (t-2)^2 = 4 + 0 + \frac{1}{2} \cdot (-4) \cdot 2^2 = -4 \mathrm{~m}\)

The particle's new coordinates at 4 seconds are \((18 \mathrm{~m}, -4 \mathrm{~m})\), showing its precise position in the two-dimensional space after evaluating both the x and y motions.