91Ó°ÊÓ

A puck is moving on an air hockey table. Relative to an \(x, y\) coordinate system at time \(t=0 \mathrm{~s},\) the \(x\) components of the puck's initial velocity and acceleration are \(v_{0 x}=+1.0 \mathrm{~m} / \mathrm{s}\) and \(a_{x}=+2.0 \mathrm{~m} / \mathrm{s}^{2} .\) The \(y\) components of the puck's initial velocity and acceleration are \(v_{0 y}=+2.0 \mathrm{~m} / \mathrm{s}\) and \(a_{y}=-2.0 \mathrm{~m} / \mathrm{s}^{2}\) Is the magnitude of the \(x\) component of the velocity increasing or decreasing in time? Is the magnitude of the \(y\) component of the velocity increasing or decreasing in time? Find the magnitude and direction of the puck's velocity at a time of \(t=0.50 \mathrm{~s}\). Specify the direction relative to the \(+x\) axis. Be sure that your calculations are consistent with your answers to the Concept Questions.

Step by step solution

01

Understanding Acceleration Effects

Determine the effect of acceleration on the velocity components. For the x-component, since the acceleration \(a_x\) is positive (\(+2.0 \, ext{m/s}^2\)), it means that the velocity in the x-direction is increasing over time. For the y-component, the acceleration \(a_y\) is negative (\(-2.0 \, ext{m/s}^2\)), meaning the velocity in the y-direction is decreasing over time.

02

Calculating the Velocity Components at \(t=0.50\) s

Use the kinematic equation \( v = v_0 + at \) for each component. For the x-component, \( v_{x} = v_{0x} + a_{x}t = 1.0 + 2.0 \times 0.5 = 2.0 \, \text{m/s} \). For the y-component, \( v_{y} = v_{0y} + a_{y}t = 2.0 - 2.0 \times 0.5 = 1.0 \, \text{m/s} \).

03

Finding the Magnitude of Velocity

The magnitude of the velocity is found using the Pythagorean theorem: \( v = \sqrt{v_{x}^{2} + v_{y}^{2}} \). Substituting our values, \( v = \sqrt{(2.0)^{2} + (1.0)^{2}} = \sqrt{4.0 + 1.0} = \sqrt{5} \approx 2.2 \, \text{m/s} \).

04

Determining the Direction of Velocity

The direction \(\theta\) can be found using the tangent function: \( \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) \). Substituting the known values, \( \theta = \tan^{-1}\left(\frac{1.0}{2.0}\right) \approx 26.6^\circ \). This angle is relative to the positive x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity Components

Velocity components refer to the parts of a vector that lie along the coordinate axes. In the context of the hockey puck example, these components determine how the puck's velocity changes along the x and y directions.

The velocity in the x-component initially is given by the equation:

• Initial x-component of velocity, \(v_{0x} = +1.0 \, \text{m/s}\)
• Acceleration in the x-direction, \(a_{x} = +2.0 \, \text{m/s}^{2}\)

The positive acceleration in the x-direction implies that the x-component velocity is increasing over time.

For the y-component:

• Initial y-component of velocity, \(v_{0y} = +2.0 \, \text{m/s}\)
• Acceleration in the y-direction, \(a_{y} = -2.0 \, \text{m/s}^{2}\)

The negative acceleration in the y-direction says the y-component of velocity is decreasing over time. This is crucial in understanding how each component behaves with respect to time.

Acceleration

Acceleration is essentially a measure of how quickly an object's velocity changes over time. It's an essential part of kinematics.

In our problem, the puck has different acceleration values in the x and y components, affecting its motion accordingly:

• In the x-component, because the acceleration \(a_{x}\) is positive, this ensures that the puck’s velocity in the x-direction continuously increases.
• Conversely, for the y-component, since the acceleration \(a_{y}\) is negative, the velocity in the y-direction decreases, showcasing different dynamics of movement along axes.

Understanding acceleration's effects on each velocity component helps us predict future motion paths and speeds. It's also key to solving kinematics problems accurately.

Magnitude of Velocity

The magnitude of velocity refers to the speed part of the velocity vector without considering its direction. It can be thought of as the hypotenuse of a right-angled triangle formed by the velocity components.

To calculate the magnitude of velocity for the puck, we use the equation: \[v = \sqrt{v_{x}^{2} + v_{y}^{2}}\] Substituting the values from our problem, we have:

• \(v_{x} = 2.0 \, \text{m/s}\) (at \(t = 0.50 \, \text{s}\))
• \(v_{y} = 1.0 \, \text{m/s}\) (at \(t = 0.50 \, \text{s}\))

Thus, the magnitude is: \[v = \sqrt{(2.0)^{2} + (1.0)^{2}} = \sqrt{4.0 + 1.0} = \sqrt{5} \approx 2.2 \, \text{m/s}\] This tells us how fast the puck is moving on the air hockey table, disregarding the direction.

Direction of Velocity

The direction of velocity indicates the angle formed by the velocity vector with a reference axis, in this case, the positive x-axis. This is crucial for understanding motion on a plane.

Using the tangent function is a common way to find this direction. Here, we use: \[ \theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}}\right) \] With the values from this exercise:

• \(v_{x} = 2.0 \, \text{m/s}\)
• \(v_{y} = 1.0 \, \text{m/s}\)

We calculate the direction: \[ \theta = \tan^{-1}\left(\frac{1.0}{2.0}\right) \approx 26.6^{\circ} \] This angle is relative to the positive x-axis and lets us know the precise direction in which the puck is headed, combining both its speed and angle of motion.