Problem 21 - Physics for Scientists and Engineers with Modern Physics

Question

At \(t=0\), a particle starts from rest at \(x=0, y=0\), and moves in the \(x y\) plane with an acceleration \(\overrightarrow{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} .\) Determine \((a)\) the \(x\) and \(y\) components of velocity, \((b)\) the speed of the particle, and \((c)\) the position of the particle, all as a function of time. \((d)\) Evaluate all the above at \(t=2.0 \mathrm{~s}\).

Step-by-Step Solution

Answer

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

Step 1: Find the Velocity Components

The particle starts from rest, so the initial velocity is 0. Using the acceleration provided, we can find the velocity components using the formula \( v(t) = a \cdot t \). For the \( x \)-component, \( v_x(t) = 4.0 \times t \). For the \( y \)-component, \( v_y(t) = 3.0 \times t \).

Step 2: Express the Speed as a Function of Time

The speed of the particle is the magnitude of the velocity vector. Thus, the speed \( v(t) \) is given by \[ v(t) = \sqrt{v_x(t)^2 + v_y(t)^2} = \sqrt{(4.0t)^2 + (3.0t)^2} = \sqrt{16.0t^2 + 9.0t^2} = 5.0t \].

Step 3: Determine the Position Components as Time Functions

To find the position components, integrate the velocity components with respect to time. So, for \( x(t) \), integrate \( v_x(t) = 4.0t \), obtaining \( x(t) = 2.0t^2 \). For \( y(t) \), integrate \( v_y(t) = 3.0t \), resulting in \( y(t) = 1.5t^2 \).

Step 4: Evaluate at t=2.0 s

Plug \( t = 2.0 \) seconds into the expressions derived: \( v_x(2) = 4.0 \times 2.0 = 8.0 \) m/s, \( v_y(2) = 3.0 \times 2.0 = 6.0 \) m/s. The speed is \( 5.0 \times 2.0 = 10.0 \) m/s, and the position is \( x(2) = 2.0 \times 2.0^2 = 8.0 \) m, \( y(2) = 1.5 \times 2.0^2 = 6.0 \) m.

Key Concepts

Velocity Components

Understanding velocity components is key to analyzing motion in the coordinate plane. A particle moving in two dimensions has its velocity broken down into two components: one along the x-axis (\( v_x \)) and another along the y-axis (\( v_y \)). These components are derived based on the particle's acceleration and time.

In our problem, the particle starts from rest, so its initial velocity is zero. The acceleration is given as \(\overrightarrow{\mathbf{a}}=(4.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}})\, \mathrm{m}/\mathrm{s}^{2}\). This tells us how fast the particle's velocity is changing in each direction over time.

• The velocity in the x-direction: To find \(v_x(t)\), multiply the x-component of acceleration \(4.0\, \mathrm{m}/\mathrm{s}^2\) with time \(t\): \(v_x(t) = 4.0t\).
• The velocity in the y-direction: Similarly, multiply the y-component \(3.0\, \mathrm{m}/\mathrm{s}^2\) with time \(t\): \(v_y(t) = 3.0t\).

As time progresses, these velocities determine how the particle's speed changes in each direction. Calculating these values helps in understanding the directional movement of the particle.

Acceleration

Acceleration tells us how quickly an object's velocity is changing. It is a vector quantity, meaning it has both magnitude and direction. In this exercise, acceleration is given as \((4.0 \hat{\mathbf{i}} + 3.0 \hat{\mathbf{j}})\).

This specific acceleration means that every second, the velocity in the x-direction increases by \(4.0 \ \mathrm{m}/\mathrm{s}\) and the velocity in the y-direction increases by \(3.0 \ \mathrm{m}/\mathrm{s}\).

• Why is this important? Because knowing the acceleration helps us predict future velocities and positions by integration over time.
• Constant acceleration like in this problem provides linear equations for velocity components, making calculations straightforward.

Understanding acceleration is crucial for solving kinematics problems, as it links directly to how objects move through space, affecting both velocity and position over time.

Position as a Function of Time

The position of a particle in motion changes over time based on its velocity function. Calculating position as a function of time involves integrating the velocity components.

For our scenario, since acceleration is constant, the simplest approach is first to determine velocity, then integrate those velocities to find the position functions.

• Position in the x-direction: Since \(v_x(t) = 4.0t\), integrate it to find \(x(t) = \int 4.0t \, dt = 2.0t^2 + C\). Given the particle starts from \(x=0\), \(C\) is zero. Thus, \(x(t) = 2.0t^2\).
• Position in the y-direction: With \(v_y(t) = 3.0t\), integrate to obtain \(y(t) = \int 3.0t \, dt = 1.5t^2 + C\). Starting from \(y=0\) means \(C\) is zero, leaving \(y(t) = 1.5t^2\).

Therefore, the position of the particle at any time \(t\) can be easily plugged into these equations to determine how far the particle has moved along the x- and y-axes, explaining its trajectory in the coordinate plane.