Problem 6 - University Physics with Modern Physics

Question

A dog running in an open field has components of velocity \(v_x\) = 2.6 m/s and \(v_y\) = -1.8 m/s at \(t_1\) = 10.0 s. For the time interval from \(t_1\) = 10.0 s to \(t_2\) = 20.0 s, the average acceleration of the dog has magnitude 0.45 m/s\(^2\) and direction 31.0\(^\circ\) measured from the +\(x\)-axis toward the +\(y\)-axis. At \(t_2\) = 20.0 s, (a) what are the \(x\)- and \(y\)-components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_1\) and \(t_2\). How do these two vectors differ?

Step-by-Step Solution

Answer

(a) \( v_x = 6.45 \, \text{m/s} \), \( v_y = 0.52 \, \text{m/s} \). (b) \( 6.47 \, \text{m/s}, \ 4.6^\circ \).

Step 1: Understanding the given information

We are given the initial velocity components as \( v_x = 2.6 \, \text{m/s} \) and \( v_y = -1.8 \, \text{m/s} \) at time \( t_1 = 10.0 \, \text{s} \). For the time interval from \( t_1 = 10.0 \, \text{s} \) to \( t_2 = 20.0 \, \text{s} \), average acceleration magnitude is \( 0.45 \, \text{m/s}^2 \) with a direction of \( 31.0^\circ \) from the +\(x\)-axis towards the +\(y\)-axis.

Step 2: Calculate acceleration components

The acceleration vector makes an angle \(31.0^\circ\) with the +\(x\)-axis. So, the \(x\)-component of the acceleration \(a_x\) is \( a_{x} = 0.45 \cos(31.0^\circ) \approx 0.385 \text{ m/s}^2 \) and the \(y\)-component \(a_y\) is \( a_{y} = 0.45 \sin(31.0^\circ) \approx 0.232 \text{ m/s}^2 \).

Step 3: Calculate the change in velocity components

Using the formula \( \Delta v = a \cdot \Delta t \), where \( \Delta t = t_2 - t_1 = 20.0 - 10.0 = 10.0 \, \text{s} \).\( \Delta v_x = a_x \cdot 10.0 = 0.385 \cdot 10.0 \approx 3.85 \, \text{m/s} \)\( \Delta v_y = a_y \cdot 10.0 = 0.232 \cdot 10.0 \approx 2.32 \, \text{m/s} \).

Step 4: Calculate the final velocity components

The final \(x\)-component of velocity is \( v_{x, f} = v_x + \Delta v_x = 2.6 + 3.85 = 6.45 \, \text{m/s} \).The final \(y\)-component of velocity is \( v_{y, f} = v_y + \Delta v_y = -1.8 + 2.32 = 0.52 \, \text{m/s} \).

Step 5: Finding the magnitude of the final velocity

The magnitude of the velocity \( v_f \) is found using Pythagorean theorem: \[ v_f = \sqrt{v_{x,f}^2 + v_{y,f}^2} = \sqrt{(6.45)^2 + (0.52)^2} \approx 6.47 \, \text{m/s} \]

Step 6: Calculating the direction of the final velocity

The direction \( \theta_f \) is given by \( \theta_f = \tan^{-1}\left(\frac{v_{y,f}}{v_{x,f}}\right) = \tan^{-1}\left(\frac{0.52}{6.45}\right) \approx 4.6^\circ \) above the +\(x\)-axis.

Step 7: Compare velocity vectors

At \( t_1 \), the velocity \( \vec{v} \) is mostly in the +\(x\) direction with a slight negative \(y\) component. At \( t_2 \), the velocity is greater in the +\(x\)-direction with a small positive \(y\) component. The main difference is the change from a slight negative to a slight positive \(y\) component, reflecting the direction of the acceleration vector being towards the +\(y\)-axis.

Key Concepts

Velocity Components

In 2D kinematics, the velocity of an object can be broken down into its component parts along the x-axis and y-axis. This is particularly useful when an object moves along a path that isn't purely linear. For instance, in our example with the dog running in an open field, the velocity at time \( t_1 \) has components \( v_x = 2.6 \, \text{m/s} \) and \( v_y = -1.8 \, \text{m/s} \). These components indicate that the dog is moving in the positive x-direction while the movement in the y-direction is slight and negative.

To find these components, you usually use trigonometry if the total velocity and angle are known. Velocity components are calculated using:

• \( v_x = v \cos(\theta) \)
• \( v_y = v \sin(\theta) \)

The x-component represents the horizontal velocity, while the y-component represents the vertical velocity. These are vital in understanding how the overall velocity changes over time, especially with the influence of acceleration.

Average Acceleration

Acceleration indicates how the velocity of an object changes over time. Average acceleration is calculated over a time interval, using the change in velocity divided by the time taken for this change. In our situation, the dog has an average acceleration of magnitude 0.45 \( \text{m/s}^2 \) over 10 seconds, in a direction 31° from the positive x-axis towards the positive y-axis.

To determine the acceleration components:

• Calculate the x-component with \( a_x = a \cos(\theta) \)
• Calculate the y-component with \( a_y = a \sin(\theta) \)

The components were calculated as \( a_x \approx 0.385 \, \text{m/s}^2 \) and \( a_y \approx 0.232 \, \text{m/s}^2 \).
Average acceleration helps us predict how the motion will evolve—your initial velocity components will be altered by the acceleration, changing the velocity components after a specific time. In this exercise, acceleration causes the dog's negative y-component of velocity to become positive by time \( t_2 \).

This explains how force (through acceleration) continuously affects motion by changing its speed and direction.

Vector Addition

Vector addition involves combining two or more vectors, which in our scenario, applies to combining velocity changes due to acceleration. When solving problems in 2D kinematics, you add velocity vectors together to determine the resulting motion.

To find the final velocity components using vector addition:

• Sum the initial x-component of velocity and the change in x-component \( v_{x,f} = v_x + \Delta v_x \)
• Sum the initial y-component of velocity and the change in y-component \( v_{y,f} = v_y + \Delta v_y \)

In this case, the final x-component is \( 6.45 \, \text{m/s} \), while the final y-component is \( 0.52 \, \text{m/s} \).

Vector addition can also help find the resultant velocity magnitude and direction:

• Magnitude is found using Pythagorean theorem: \( v_f = \sqrt{v_{x,f}^2 + v_{y,f}^2} \)
• Direction is given by \( \theta_f = \tan^{-1}(\frac{v_{y,f}}{v_{x,f}}) \)

These calculations not only help find the final movement but offer insights into how the path evolves over time. The dog's velocity vector at \( t_2 \) lays mostly in the positive x-direction with a shift in the y-direction, showcasing the overall motion change.