Problem 24 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A particle starts from the origin at \(t=0\) with an initial velocity of \(5.0 \mathrm{~m} / \mathrm{s}\) along the positive \(x\) axis. If the acceleration is \((-3.0 \hat{\mathbf{i}}+4.5 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2},\) determine the velocity and position of the particle at the moment it reaches its maximum \(x\) coordinate.

Step-by-Step Solution

Answer

The velocity is \(7.515 \hat{\mathbf{j}}\) m/s and the position is \(4.18 \hat{\mathbf{i}} + 6.268 \hat{\mathbf{j}}\) m.

Step 1: Determine the condition at maximum x-coordinate

The maximum x-coordinate is reached when the velocity component along the x-axis is zero. Given initial velocity in x is 5.0 m/s and the acceleration in x is -3.0 m/s², we need to find the time when velocity in x equals zero for the condition of maximum x-coordinate.

Step 2: Find time when velocity in x is zero

The velocity in the x-direction is given by \(v_x = v_{0x} + a_x t\). Here, \(v_{0x} = 5.0 \text{ m/s}\) and \(a_x = -3.0 \text{ m/s}^2\).Set \(v_x = 0\):\[0 = 5.0 - 3.0t\]Solving for \(t\):\[t = \frac{5.0}{3.0} \approx 1.67 \text{ s}\]

Step 3: Calculate the velocity at this time

Using the time \(t = 1.67 \text{ s}\), calculate the velocity in y-direction. The initial velocity in y is zero, and acceleration in y is 4.5 m/s².\(v_y = v_{0y} + a_y t\)\(v_y = 0 + 4.5 \times 1.67\)\(v_y = 7.515 \text{ m/s}\)Thus the velocity of the particle, \(\mathbf{v} = 0 \hat{\mathbf{i}} + 7.515 \hat{\mathbf{j}}\).

Step 4: Calculate the position at this time

To find the position, calculate \(x\) and \(y\) coordinates separately. For \(x\):\(x = v_{0x}t + \frac{1}{2}a_xt^2\)\(x = 5.0 \times 1.67 + \frac{1}{2}(-3.0)(1.67)^2\)\(x = 8.35 - 4.17 \approx 4.18 \text{ m}\)For \(y\):\(y = v_{0y}t + \frac{1}{2}a_yt^2\)\(y = 0 + \frac{1}{2}(4.5)(1.67)^2\)\(y = 6.268 \text{ m}\)Thus the position is, \(\mathbf{r} = 4.18 \hat{\mathbf{i}} + 6.268 \hat{\mathbf{j}}\).

Step 5: Conclusion

The velocity of the particle at maximum x is \(\mathbf{v} = 7.515 \hat{\mathbf{j}}\) and the position is \(\mathbf{r} = 4.18 \hat{\mathbf{i}} + 6.268 \hat{\mathbf{j}}\).

Key Concepts

Projectile Motion

Projectile motion is a key concept in kinematics that deals with objects that are launched into the air and subject only to gravity and/or other constant accelerations. This kind of motion is seen often in real-life situations, such as a football being kicked or a cannonball being fired. In this exercise, the motion is primarily influenced by the acceleration components in two directions: horizontal (x-axis) and vertical (y-axis). The initial setup defines a particle starting with a particular velocity and subjected to a constant acceleration.

• Initial Velocity: The particle begins its motion with an initial velocity of 5.0 m/s along the x-axis and 0 m/s along the y-axis.
• Acceleration: The particle experiences a constant acceleration of (-3.0 m/s²) along the x-axis and (4.5 m/s²) along the y-axis.

The challenge is to find when the particle reaches a maximum x-coordinate. This occurs when the velocity component on the x-axis becomes zero, indicating a change in direction. By evaluating the time when this happens, we can determine both the position and velocity of the particle at that moment.

Vector Calculations

In kinematics, vector calculations are crucial for solving problems involving multiple dimensions, as they allow us to handle different direction components separately. In this scenario, we have a two-dimensional vector problem that involves decomposing and analyzing vectors along the x and y axes.
Twe crucial components:

• X-axis Calculations: Initially, the particle has a velocity of 5.0 m/s. The acceleration on the x-axis is negative (-3.0 m/s²), causing a reduction in the velocity as time progresses. By setting the final velocity to zero, we can compute when the particle stops moving forward along the x-axis.
• Y-axis Calculations: The acceleration along the y-axis is 4.5 m/s². With no initial velocity, this acceleration directly increases the velocity over time. Using vector equations, we calculate the change in velocity over specific time intervals.

These calculations help us identify how the particle's velocity and position evolve over time, providing insights needed to solve the problem. By analyzing these vector components independently and then combining their effects, we achieve a full understanding of the projectile motion.

Acceleration

Acceleration plays a fundamental role in determining the motion of any object, influencing both velocity and position. It is defined as the rate of change of velocity with respect to time. In this exercise, it is essential for predicting the motion of the particle along both axes.
There are a few critical elements to consider:

• Negative Acceleration on X-axis: The acceleration of -3.0 m/s² means the particle is slowing down as it moves along the x-axis. Eventually, it comes to a stop momentarily, creating a turning point or a maximum x-position.
• Positive Acceleration on Y-axis: The positive 4.5 m/s² acceleration means the particle's velocity in the y-direction increases over time. There is no initial upward motion, so this positive acceleration is the only factor driving the movement along the y-axis.

Understanding the effects of each acceleration component is crucial for determining how the particle moves, changes velocity, and the resulting trajectory it follows.