91Ó°ÊÓ

Understanding particle motion is fundamental in physics, especially in dynamic systems where objects move along specific paths. Here, we are examining a particle moving along a curve described by the equation \( y = 2 \sin(2x) \). This means the particle's path is not linear, but follows a wave-like pattern determined by the sine function.

The particle has an x-component of velocity that remains constant at 2 m/s. This constant velocity implies uniform motion along the x-axis. Such uniform motion simplifies calculations, as the position can be determined directly by multiplying velocity by time. In contrast, motion in the y-direction relies on the curve's parametric form, which is inherently more complex.

It's essential to understand that motion in physics often involves multiple components, such as the x and y directions in this problem. These components can be analyzed separately and then combined to find the resultant motion or displacement.

Parametric equations are a powerful tool used to describe paths or curves in terms of a separate parameter, often time \( t \). In this exercise, the behavior of the particle is described by both \( x(t) \) and \( y(t) \).

Here:

• The x-component equation is defined through velocity: \( x = 2t \).
• The y-component uses the sinusoidal function: \( y(t) = 2 \sin(4t) \).

These equations help map out the path of the particle over time. By substituting the parameter, we can solve the equations to find specific positions of the particle at any given time. This approach allows for a comprehensive understanding of motion in two dimensions.

One of the strengths of parametric equations is that they allow you to independently evaluate the separate dimensions of motion. In this scenario, once \( t \) is found using one equation, the calculated value is applied in another, linking the two dimensions of movement.

Displacement is the overall change in position of a particle after a journey, taking into account both directions. In this scenario, the displacement is a vector with both an x-component and a y-component.

For the x-direction, since velocity is constant, displacement simply involves multiplying the velocity (2 m/s) by the total travel time (\( t = \frac{\pi}{4} \) seconds). Thus, the x-displacement is \( \frac{\pi}{2} \) meters.

The y-direction involves integration due to the dynamic nature of the path. The equation \( y(t) = 2 \sin(4t) \) is integrated over the interval from 0 to \( t = \frac{\pi}{4} \) to get the y-displacement. This results in a y-displacement of \( \frac{1}{2} \) meters.

The magnitude of the total displacement vector combines these two components by applying the Pythagorean theorem: \[ d = \sqrt{(s_x)^2 + (s_y)^2} \], where \( s_x \) and \( s_y \) are displacements in the x and y directions respectively.

Trigonometric integration is crucial for calculating areas and solving equations that involve trigonometric functions. When a particle's path is defined by such functions, integration allows us to find specific displacements over time.

In this problem, the function \( y(t) = 2 \sin(4t) \) describes the vertical position of the particle. To find how much the particle moves vertically from the origin to \( t = \frac{\pi}{4} \), one needs to integrate this function over the specified interval. The integration:\[ \int_0^{\pi/4} 2 \sin(4t) \, dt \]results in finding the total change in the y-direction over the given time.

Through the process, the integral solution represents the accumulated "area" under the y-function, giving us the desired displacement. Trigonometric integrals often simplify to neat numerical values, which are easier to work with and compare against potential answer choices.