91Ó°ÊÓ

In the motion of particles, the position vector plays a crucial role. It helps us determine the location of an object in a coordinate system, typically described in terms of x and y coordinates in two dimensions. For a particle moving in the x-y plane, the position vector \( \vec{r} \) is often given by the expression \( \vec{r} = x(t) \hat{i} + y(t) \hat{j} \), where \( x(t) \) and \( y(t) \) represent the particle's position at any time \( t \).

In the provided problem, the position vector is expressed as \( \vec{r} = 2t \hat{i} + 2 \sin\left(\frac{\pi}{4} t\right) \hat{j} \). This vector gives us functions for both the x-component \( x(t) = 2t \) and the y-component \( y(t) = 2\sin\left(\frac{\pi}{4} t\right) \).
By calculating these functions at different time intervals \( t = 0, 1.0, 2.0, 3.0, 4.0 \) seconds, we can sketch the path of the particle. This helps visualize how it moves over time in the plane.

• At \( t = 0 \): \( x = 0 \), \( y = 0 \).
• At \( t = 1.0 \): \( x = 2 \), \( y \approx 1.41 \).
• At \( t = 2.0 \): \( x = 4 \), \( y = 2 \).
• At \( t = 3.0 \): \( x = 6 \), \( y \approx 1.41 \).
• At \( t = 4.0 \): \( x = 8 \), \( y = 0 \).

Plotting these points provides a visual trajectory of the particle's motion.

The concept of velocity in physics refers to the rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude and direction. Velocity is derived by differentiating the position vector with respect to time. Thus, for a position vector \( \vec{r} \), the velocity \( \vec{v} \) is given by \( \vec{v} = \frac{d\vec{r}}{dt} \).
In our exercise, when we differentiate the position vector \( \vec{r} = 2t \hat{i} + 2 \sin\left(\frac{\pi}{4} t\right) \hat{j} \), we find the velocity \( \vec{v} = 2 \hat{i} + \frac{\pi}{2} \cos\left(\frac{\pi}{4} t\right) \hat{j} \). This expression gives the components of velocity at any time \( t \).

• At \( t = 1.0 \): \( v_x = 2 \), \( v_y \approx 1.11 \).
• At \( t = 2.0 \): \( v_x = 2 \), \( v_y = 0 \).
• At \( t = 3.0 \): \( v_x = 2 \), \( v_y \approx -1.11 \).

This indicates how fast and in what direction the particle is moving at these times. By plotting the velocity vectors on the path outlined in the position section, you can see that the velocity is always tangent to the path, confirming the direction of motion.

Acceleration in physics describes how the velocity of an object changes with time. Like velocity, it is also a vector quantity and is obtained by differentiating the velocity vector with respect to time. In essence, acceleration tells us about the rate of change of velocity for a moving object.
In the given problem, to find the acceleration \( \vec{a} \) of the particle, we differentiate the velocity expression \( \vec{v} = 2 \hat{i} + \frac{\pi}{2} \cos\left(\frac{\pi}{4} t\right) \hat{j} \) to get \( \vec{a} = 0 \hat{i} - \frac{\pi^2}{8} \sin\left(\frac{\pi}{4} t\right) \hat{j} \).
Evaluating these components at specific times:

• At \( t = 1.0 \): \( a_x = 0 \), \( a_y \approx -0.87 \).
• At \( t = 2.0 \): \( a_x = 0 \), \( a_y \approx -1.23 \).
• At \( t = 3.0 \): \( a_x = 0 \), \( a_y \approx -0.87 \).

This demonstrates that while the particle moves forward consistently along the x-direction with constant velocity, the y-component changes sinusoidally, resulting in a varying acceleration in the y-direction. Understanding how acceleration impacts motion is key in predicting future positions of the particle in dynamic systems.