91Ó°ÊÓ

Velocity is a key concept in kinematics and describes the rate of change of an object's position. It is a vector quantity, meaning it has both magnitude and direction. In our problem, the initial velocity of the particle was given as \( \vec{v} = (3.00 \hat{\mathrm{i}}) \mathrm{m/s} \), meaning it moves in the positive x-direction with a speed of 3.00 m/s.

As the particle experiences a constant acceleration, its velocity changes over time. Recall that velocity along any axis can be calculated using the equation:

• \( v_x = v_{0x} + a_x t \)

This captures the x-component of velocity where \( v_{0x} \) is the initial velocity, \( a_x \) is the acceleration, and \( t \) is time. As the particle reaches its maximum x-coordinate, its velocity along the x-axis becomes zero. This happens because the acceleration affects the velocity negatively in this direction.

The y-component of velocity at this time was calculated as \( v_y = -1.50 \mathrm{m/s} \), leading to the velocity vector \( \vec{v} = 0 \hat{\mathrm{i}} - 1.50 \hat{\mathrm{j}} \) at the maximum x-coordinate point. So, the particle is moving downward.

Acceleration is another fundamental concept in kinematics, describing the rate of change of velocity with respect to time. It is also a vector, highlighting both magnitude and direction. In this scenario, the particle's acceleration was given as \( \vec{a} = (-1.00 \hat{\mathrm{i}} - 0.500 \hat{\mathrm{j}}) \mathrm{m/s^2} \).

This means that the particle is experiencing a deceleration in the x-direction (as indicated by the negative sign in the x-component) and a constant acceleration downward in the y-direction.

To determine changes in velocity over time:

• Use the equation \( v = v_0 + at \)

When discussing acceleration, it's crucial to note that it directly influences the time it takes for the velocity to change and the subsequent trajectory of the particle.

In our problem, the effect of acceleration is evident in how the particle reaches a maximum x-coordinate when its x-component of velocity reduces to zero. This occurs due to the deceleration in the x-direction, aligning with the key idea that acceleration influences velocity and, ultimately, the position.

The position vector tells us the location of an object at any given time by providing the distance from a reference point (often the origin) in space. In this exercise, we see the use of the position vector formula:

• \( \vec{r} = \vec{r}_{0} + \vec{v}_{0}t + \frac{1}{2}\vec{a}t^2 \)

This factors in the initial position, initial velocity, time, and acceleration to find the new position.

At the maximum x-coordinate, the time was calculated to be 3.00 seconds. Plugging this value into the position vector equation helps us compute the x and y positions separately. The x-coordinate is found to be 4.50 meters, while the y-coordinate is \(-2.25\) meters. This yields the position vector:

• \( \vec{r} = 4.50 \hat{\mathrm{i}} - 2.25 \hat{\mathrm{j}} \)

Through the position vector, we see that the particle has moved horizontally and dropped vertically, showcasing how velocity and acceleration impact its overall trajectory.