91Ó°ÊÓ

When we talk about velocity components in kinematics, we're discussing how the velocity of an object can be split into two parts based on directions, typically along the x-axis and the y-axis. This helps us understand how fast the object is moving horizontally and vertically. For the remote-controlled car exercise, the velocity is given by two parts:

• The x-component: \(v_x(t) = 5.00 - 0.0180t^2\).
• The y-component: \(v_y(t) = 2.00 + 0.550t\).

The first step to solving these problems usually involves differentiating to find the acceleration components, but understanding these initial velocity components is crucial.
At any given time, say at \(t = 8 \text{s}\), you plug in the time value to find out how fast the car is moving horizontally (x-direction) and vertically (y-direction).
So, at \(t = 8 \text{s}\), the velocity components of the car are:

• Horizontal: \(v_x(8) = 5.00 - 0.0180 \times 8^2 = 3.85 \text{ m/s}\)
• Vertical: \(v_y(8) = 2.00 + 0.550 \times 8 = 6.40 \text{ m/s}\)

Recognizing these velocity components gives us a good picture of how the car behaves in terms of speed in both horizontal and vertical directions.

Acceleration components tell us how the velocity of the car changes over time in each direction—horizontally and vertically. To find these, we differentiate the velocity expressions. For our remote-controlled car:

• Initially, we calculate the horizontal acceleration: \(a_x(t) = \frac{d}{dt}(5.00 - 0.0180t^2) = -0.0360t\).
• The vertical acceleration is simply the derivative of the constant slope in \(v_y(t)\): \(a_y(t) = \frac{d}{dt}(2.00 + 0.550t) = 0.550\).

This means that the horizontal component of the car's velocity is decreasing over time due to a negative acceleration factor. Contrarily, the vertical component is constantly increasing since the acceleration is positive and steady.
At \(t = 8 \text{s}\), the car's acceleration is:

• Horizontal: \(a_x(8) = -0.0360 \times 8 = -0.288 \text{ m/s}^2\)
• Vertical: \(a_y = 0.550 \text{ m/s}^2\)

These components help determine how quickly and in what manner the car changes its velocity as it moves across the parking lot.

In kinematics, knowing the magnitude and direction of vectors like velocity and acceleration gives a complete picture of motion. The magnitude of a vector gives us an idea of how strong the vector is, whereas the direction tells us the way it points.
To find the magnitude of velocity at any time, use:\[|\vec{v}| = \sqrt{v_x^2 + v_y^2}\]At \(t = 8 \text{s}\), the results are:

• \(|\vec{v}| = \sqrt{(3.85)^2 + (6.40)^2} \approx 7.47 \text{ m/s}\).

Meanwhile, its direction is found with the arctangent function:\[\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\]Applying the data for \(t = 8 \text{s}\), we find:

• The direction is \(\theta \approx 58.3^\circ\).

For acceleration, follow a similar method:\[|\vec{a}| = \sqrt{a_x^2 + a_y^2}\]And find for the given time as:

• \(|\vec{a}| \approx 0.620 \text{ m/s}^2\).
• Direction: \(\theta = \tan^{-1}\left(\frac{0.550}{-0.288}\right) \approx -62.4^\circ\).

The negative angle denotes direction below the horizontal line in standard position, effectively illustrating acceleration vector direction.