Problem 89 - Fundamentals of Physics

Question

A particle starts from the origin at \(t=0\) with a velocity of \(8.0 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\) and moves in the \(x y\) plane with constant acceleration \((4.0 \hat{\mathrm{i}}+2.0 \mathrm{j}) \mathrm{m} / \mathrm{s}^{2} .\) When the particle's \(x\) coordinate is \(29 \mathrm{~m}\), what are its (a) \(y\) coordinate and (b) speed?

Step-by-Step Solution

Answer

The y coordinate is 48.2 m, and the speed is 21.80 m/s.

Step 1: Identify and Write Down Known Values

The initial velocity of the particle is given by \( \vec{u} = 8.0 \hat{\mathrm{j}} \mathrm{m/s} \). The acceleration is given by \( \vec{a} = 4.0 \hat{\mathrm{i}} + 2.0 \hat{\mathrm{j}} \mathrm{m/s^2} \). The condition is that the \( x \) coordinate reaches 29 m. We need to find the corresponding \( y \) coordinate and the speed at this position.

Step 2: Use Kinematic Equation for x Component

Use the equation \( x = u_x t + \frac{1}{2} a_x t^2 \). Here, \( u_x = 0 \) since the initial velocity has no \( i \) component. The equation simplifies to \( x = \frac{1}{2} a_x t^2 \). Plugging in the values, we get \( 29 = \frac{1}{2} \times 4.0 \times t^2 \), which simplifies to \( t^2 = \frac{29}{2} \), resulting in \( t = \sqrt{\frac{29}{2}} \approx 3.81 \, \text{s} \).

Step 3: Use Kinematic Equation for y Component

Substitute the time found into the y-component equation \( y = u_y t + \frac{1}{2} a_y t^2 \). Here, \( u_y = 8.0 \) and \( a_y = 2.0 \). Thus, \( y = 8.0 \times 3.81 + \frac{1}{2} \times 2.0 \times (3.81)^2 \), which calculates to \( y \approx 40.94 + 7.26 = 48.2 \, \text{m} \).

Step 4: Calculate the Final Velocity Vectors

Using \( v_x = u_x + a_x t \) and \( v_y = u_y + a_y t \), where \( u_x = 0 \), \( v_x = 0 + 4.0 \times 3.81 = 15.24 \, \text{m/s} \). Similarly, \( v_y = 8.0 + 2.0 \times 3.81 \), giving \( v_y = 15.62 \, \text{m/s} \).

Step 5: Calculate Speed of the Particle

The speed \( v \) of the particle is the magnitude of the velocity vector: \( v = \sqrt{v_x^2 + v_y^2} \). Substituting the values: \( v = \sqrt{15.24^2 + 15.62^2} \approx \sqrt{475.50} \), resulting in \( v \approx 21.80 \, \text{m/s} \).

Key Concepts

Kinematic Equations

Kinematic equations are a set of equations that describe the motion of objects in terms of their velocity, acceleration, and displacement over time. In kinematics, we often focus on rectilinear motion, or motion along a straight line, however, the principles apply to two or three dimensions as well.
These equations are invaluable, allowing us to solve problems such as the starting exercise, where we explore the motion of a particle with known initial conditions and full acceleration equations.

The fundamental set of kinematic equations include:

• For velocity: \( v = u + at \)
• For displacement: \( s = ut + \frac{1}{2}at^2 \)
• Another form of displacement: \( s = vt - \frac{1}{2}at^2 \)
• To link velocity, displacement, and acceleration: \( v^2 = u^2 + 2as \)

These equations assume constant acceleration and can be adjusted for varying dimensions by considering vector components. This was key in solving the problem as the particle moved in the xy-plane.

Velocity

Velocity is a vector quantity that describes the speed and direction of a moving object. The word "vector" is crucial here because it means we're dealing not just with how fast something moves, but also the direction it's heading.
In our exercise, the initial velocity had a component only in the j (or y) direction, specifically \( 8.0 \hat{\mathrm{j}} \ \mathrm{m/s} \).
This tells us that the particle began its movement directly upward relative to our fixed coordinate system.

Velocity components can be separated into the following:

• The initial velocity \( u \), denoted by components \( u_x \) and \( u_y \).
• The final velocity \( v \) after a time \( t \), given by \( v_x = u_x + a_x t \) and \( v_y = u_y + a_y t \).

This understanding enabled us to ascertain not only where the particle was moving but how fast in each directional axis, providing essential insight for our calculations.

Acceleration

Acceleration is another vector quantity and measures the rate of change of velocity. In the problem, acceleration was constant and described by the vector \( \vec{a} = 4.0 \hat{\mathrm{i}} + 2.0 \hat{\mathrm{j}} \ \mathrm{m/s^2} \). This vector tells us that for every second, the velocity in the x-direction increases by 4 m/s, and that in the y-direction increases by 2 m/s.

Key aspects of acceleration include:

• Influence on velocity: It changes both the speed and direction of movement.
• Effect on displacement: Given by the equation \( s = ut + \frac{1}{2}at^2 \), acceleration plays a significant role in determining how far and in which direction an object will move.
• Combination with initial conditions: The initial velocity vector and acceleration vector together predict future states of motion.

This relationship is pivotal in kinematics, offering a complete framework to predict a particle's motion in different scenarios.

Coordinate System

The coordinate system is a crucial setting for any kinematic problem. It provides a background convention for defining positions, velocities, and accelerations.
In the xy-plane specified in our task, the system allows us to break down vectors into their components — typically along the x and y axes.

This division into orthogonal components (x and y) simplifies complex two-dimensional motion into two separate one-dimensional problems:

• The x-axis motion, completely described by its own set of equations for displacement, velocity, and acceleration.
• The y-axis motion, equally governed by similar independent equations.

By solving each axis independently, we can then reconstruct the overall motion of an object, providing a detailed map of its path.
This particular setup is essential for clear understanding and problem-solving, enabling us to calculate precisely where the particle ends up and how fast it was moving when it reached specific coordinates.