Problem 21 - Physics for Scientists and Engineers with Modern Physics

Question

(II) At \(t=0,\) a particle starts from rest at \(x=0, y=0\) , and moves in the \(x y\) planc with an acceleration $$ \begin{array}{l}{\vec{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} . \text { Determine }(a) \text { the } x \text { and } y \text { compo- }} \\ {\text { nents of velocity, }(b) \text { the spced of the particle, and }(c) \text { the }}\end{array} $$ position of the particle, all as a function of time. (d) Eval- uate all the above at \(t=2.0 \mathrm{s}\) .

Step-by-Step Solution

Answer

At \(t = 2.0\, \text{s}\), \(v_x = 8.0\, \text{m/s}\), \(v_y = 6.0\, \text{m/s}\), speed is 10.0 m/s, position is \((8.0, 6.0)\, \text{m}\).

Step 1: Find the Velocity Components

The velocity is the integral of acceleration with respect to time. Given the constant accelerations for x and y components as \( a_x = 4.0 \) and \( a_y = 3.0 \), we integrate to find velocity.\( v_x(t) = \int a_x \, dt = \int 4.0 \, dt = 4.0t + C_x \)\( v_y(t) = \int a_y \, dt = \int 3.0 \, dt = 3.0t + C_y \)Since the particle starts from rest, \( C_x = 0 \) and \( C_y = 0 \), giving:\( v_x(t) = 4.0t \)\( v_y(t) = 3.0t \)

Step 2: Calculate the Speed of the Particle

The speed is the magnitude of the velocity vector. \( |v(t)| = \sqrt{v_x^2 + v_y^2} = \sqrt{(4.0t)^2 + (3.0t)^2} = \sqrt{16.0t^2 + 9.0t^2} = \sqrt{25.0t^2} = 5.0t \)

Step 3: Determine the Position Components

The position is the integral of velocity with respect to time.For x-position:\( x(t) = \int v_x(t) \, dt = \int 4.0t \, dt = 2.0t^2 + C_x' \)For y-position:\( y(t) = \int v_y(t) \, dt = \int 3.0t \, dt = 1.5t^2 + C_y' \)Position at \( t = 0 \) is \( (0, 0) \), so \( C_x' = 0 \) and \( C_y' = 0 \), giving:\( x(t) = 2.0t^2 \)\( y(t) = 1.5t^2 \)

Step 4: Evaluate at t=2.0 s

Substitute \( t = 2.0 \) into the equations obtained.Velocity:\( v_x(2.0) = 4.0 \times 2.0 = 8.0 \, \text{m/s} \)\( v_y(2.0) = 3.0 \times 2.0 = 6.0 \, \text{m/s} \)Speed:\( |v(2.0)| = 5.0 \times 2.0 = 10.0 \, \text{m/s} \)Position:\( x(2.0) = 2.0 \times (2.0)^2 = 8.0 \, \text{m} \)\( y(2.0) = 1.5 \times (2.0)^2 = 6.0 \, \text{m} \)

Key Concepts

Acceleration

Acceleration refers to the rate at which an object changes its velocity. In our exercise, we see the particle having a constant acceleration in two directions: the x-axis and the y-axis. Specifically, the acceleration vector is given by \( \vec{\mathbf{a}} = (4.0 \hat{\mathbf{i}} + 3.0 \hat{\mathbf{j}}) \mathrm{m/s}^{2} \). This means that every second, the velocity in the x-direction increases by 4.0 m/s, while in the y-direction, it increases by 3.0 m/s.

Since acceleration is constant, finding the velocity involves integrating the acceleration function with respect to time. For the x-component of acceleration, \( a_x = 4.0 \ \mathrm{m/s}^{2} \), the velocity after integrating becomes \( v_x(t) = 4.0t \). Similarly, for the y-component, \( a_y = 3.0 \ \mathrm{m/s}^{2} \), the velocity is \( v_y(t) = 3.0t \).

An important thing to remember is that when a particle starts from rest, its initial velocity is zero. This fact allows us to simplify our velocity expressions by confirming that the constants of integration are zero.

Velocity

Velocity is a vector quantity that describes the speed of an object along with its direction. In kinematics, it is crucial to determine the velocity to understand how the position of an object changes over time. Velocity has both magnitude and direction, and in two dimensions, it is split into components that are easier to analyze separately.

From our exercise, the velocity components are derived by integrating the acceleration. We have \( v_x(t) = 4.0t \) for the x-direction and \( v_y(t) = 3.0t \) for the y-direction. These equations show how the velocity of the particle changes linearly with time due to constant acceleration.

The speed of the particle, which is the magnitude of the velocity vector, is calculated using Pythagorean theorem: \( |v(t)| = \sqrt{v_x^2 + v_y^2} = 5.0t \). This expression gives us insight into how fast the particle is moving at any time \( t \). Velocity, unlike speed, always includes direction, which is important when solving two-dimensional motion problems.

Position

Position refers to the specific location of an object in space at any given time. In kinematics, it is essential to calculate the position to fully describe the motion of a particle. For our problem, we start with the particle being at the origin, \( (x, y) = (0, 0) \), at \( t = 0 \). This initial condition simplifies further calculations as it means our constants of integration are zero.

By integrating the velocity with respect to time, we determine the position components. For the x-direction, integrating \( v_x(t) = 4.0t \) gives \( x(t) = 2.0t^2 \). Similarly, integrating \( v_y(t) = 3.0t \) yields \( y(t) = 1.5t^2 \).

These equations describe how the position of the particle changes over time. At \( t = 2.0s \), we can determine the position as \( x(2.0) = 8.0 \ \text{m} \) and \( y(2.0) = 6.0 \ \text{m} \). This demonstrates the particle's movement within the two-dimensional plane, providing a complete picture of its trajectory over time.