Problem 87 - Physics for Scientists and Engineers with Modern Physics

Question

A particle has a velocity of \(\vec{\mathbf{v}}=(-2.0 \hat{\mathbf{i}}+3.5 t \mathbf{j}) \mathrm{m} / \mathrm{s}\) . The particle starts at \(\vec{\mathbf{r}}=(1.5 \hat{\mathrm{i}}-3.1 \hat{\mathrm{j}}) \mathrm{m}\) at \(t=0 .\) Give the position and acceleration as a function of time. What is the shape of the resulting path?

Step-by-Step Solution

Answer

The path is a parabolic trajectory, given by \( \vec{\mathbf{r}}(t) = (1.5 - 2.0 t) \hat{\mathbf{i}} + (1.75 t^2 - 3.1) \hat{\mathbf{j}} \).

Step 1: Find the acceleration

The velocity of the particle is given as a function of time: \( \vec{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \text{ m/s} \). To find the acceleration, differentiate the velocity with respect to time. The derivative of the velocity gives us: \( \frac{d\vec{\mathbf{v}}}{dt} = \vec{\mathbf{a}} = (0 \hat{\mathbf{i}} + 3.5 \hat{\mathbf{j}}) \text{ m/s}^2 \). Thus, the acceleration is constant and only in the \(\hat{\mathbf{j}}\) direction.

Step 2: Set up the position-time relation

To get the position as a function of time, we integrate the velocity function with respect to time. The initial position is given as \( \vec{\mathbf{r}}_0 = (1.5 \hat{\mathbf{i}} - 3.1 \hat{\mathbf{j}}) \text{ m} \). Integration gives: \[ \vec{\mathbf{r}}(t) = \int \vec{\mathbf{v}}(t) \, dt = \int (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \, dt \]. This integrates to: \( \vec{\mathbf{r}}(t) = (-2.0 t + C_i) \hat{\mathbf{i}} + (1.75 t^2 + C_j) \hat{\mathbf{j}} \).

Step 3: Apply initial conditions

To find the constants \( C_i \) and \( C_j \), we apply the initial condition \( \vec{\mathbf{r}}(0) = (1.5 \hat{\mathbf{i}} - 3.1 \hat{\mathbf{j}}) \text{ m} \). At \( t = 0 \), we have: \( \vec{\mathbf{r}}(0) = C_i \hat{\mathbf{i}} + C_j \hat{\mathbf{j}} \). So, \( C_i = 1.5 \) and \( C_j = -3.1 \). Substituting these in gives us the position function: \( \vec{\mathbf{r}}(t) = (1.5 - 2.0 t) \hat{\mathbf{i}} + (1.75 t^2 - 3.1) \hat{\mathbf{j}} \).

Step 4: Determine the path shape

Examine the equation of the position vector. We have \( x(t) = 1.5 - 2.0 t \) for the \( \hat{\mathbf{i}} \) component and \( y(t) = 1.75 t^2 - 3.1 \) for the \( \hat{\mathbf{j}} \) component. The \( x(t) \) component is linear, while the \( y(t) \) component is quadratic, indicating that the trajectory is a parabola as it opens in the \( \mathbf{j} \) direction.

Key Concepts

Acceleration

In the study of kinematics, acceleration is a key concept that describes how the velocity of a particle changes with time.
In our exercise, the velocity is given as a time-dependent vector function \( \vec{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \text{ m/s} \).
To find the acceleration, we differentiate this velocity function with respect to time.

• The differentiation results in \( \frac{d\vec{\mathbf{v}}}{dt} = \vec{\mathbf{a}} = (0 \hat{\mathbf{i}} + 3.5 \hat{\mathbf{j}}) \text{ m/s}^2 \).
• This tells us that the acceleration here is constant.
• Notably, the acceleration has no component in the \( \hat{\mathbf{i}} \) direction, meaning the movement changes only along the \( \hat{\mathbf{j}} \) axis.

This constant acceleration results in the velocity changing uniformly along the vertical direction, shaping the particle's trajectory.

Velocity

Velocity in kinematics represents the rate of change of position with respect to time.
It is a vector, so it has both magnitude and direction. In the exercise provided, the particle's velocity is given by:

• \( \vec{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \text{ m/s} \).
• The \( -2.0 \hat{\mathbf{i}} \) component indicates a constant motion to the left along the horizontal plane (\( \hat{\mathbf{i}} \)).
• The \( 3.5 t \hat{\mathbf{j}} \) component indicates an increasing motion upwards as time progresses (\( \hat{\mathbf{j}} \)).

The velocity function hints that the particle moves linearly with time horizontally, while vertically, it speeds up or slows down depending on the time variable \( t \). This varying velocity in the \( \hat{\mathbf{j}} \) direction is driven by the acceleration, leading to a parabolic trajectory.

Position as a Function of Time

The position as a function of time in kinematics shows where a particle is located at any given time. To find this, the velocity function must be integrated over time.
From the exercise:

• The initial position \( \vec{\mathbf{r}}_0 \) is given as \( (1.5 \hat{\mathrm{i}} - 3.1 \hat{\mathrm{j}}) \text{ m} \).
• Integrating the velocity function \( \vec{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \text{ m/s} \) results in the position function: \( \vec{\mathbf{r}}(t) = (1.5 - 2.0 t) \hat{\mathbf{i}} + (1.75 t^2 - 3.1) \hat{\mathbf{j}} \).

Adding the initial conditions allows us to determine any constants from integration, ensuring the correct position at \( t=0 \).
The position function reveals a linear horizontal movement and a quadratic vertical movement, together describing a parabolic path of the particle.