Problem 87 - Physics for Scientists and Engineers with Modern Physics

Question

A particle has a velocity of \(\overrightarrow{\mathbf{v}}=(-2.0 \hat{\mathbf{i}}+3.5 t \mathbf{j}) \mathrm{m} / \mathrm{s}\). The particle starts at \(\overrightarrow{\mathbf{r}}=(1.5 \hat{\mathbf{i}}-3.1 \hat{\mathbf{j}}) \mathrm{m}\) 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 position is \( \overrightarrow{\mathbf{r}}(t) = (1.5 - 2.0t)\hat{\mathbf{i}} + (-3.1 + 1.75t^2)\hat{\mathbf{j}} \); acceleration is \( \overrightarrow{\mathbf{a}} = 3.5\hat{\mathbf{j}} \); the path is parabolic.

Step 1: Identify the Velocity Function

The velocity of the particle is given by \( \overrightarrow{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \, \text{m/s} \). This expression tells us how the velocity changes with time in terms of components along the \( \hat{\mathbf{i}} \) and \( \hat{\mathbf{j}} \) directions.

Step 2: Compute the Position Function

The position function is obtained by integrating the velocity function with respect to time, \( t \). We have:\[\begin{align*}\overrightarrow{\mathbf{r}}(t) &= \int \overrightarrow{\mathbf{v}}(t) \, dt + \overrightarrow{\mathbf{r}_0}, \ &= \int (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}} ) \, dt + (1.5 \hat{\mathbf{i}} - 3.1 \hat{\mathbf{j}})\end{align*}\]Integrating:\[\begin{align*}&= \left(-2.0 t \hat{\mathbf{i}} + \frac{3.5}{2} t^2 \hat{\mathbf{j}} \right) + (1.5 \hat{\mathbf{i}} - 3.1 \hat{\mathbf{j}}),\end{align*}\]so the position is:\[\overrightarrow{\mathbf{r}}(t) = (1.5 - 2.0 t )\hat{\mathbf{i}} + \left( -3.1 + \frac{3.5}{2} t^2 \right) \hat{\mathbf{j}}.\]

Step 3: Calculate Acceleration Function

Acceleration is the time derivative of the velocity function. Differentiate \( \overrightarrow{\mathbf{v}} \) with respect to \( t \):\[\overrightarrow{\mathbf{a}} = \frac{d}{dt} (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) = (0 \hat{\mathbf{i}} + 3.5 \hat{\mathbf{j}}) \, \text{m/s}^2.\]

Step 4: Determine the Shape of the Path

The position components are \( x(t) = 1.5 - 2.0 t \) and \( y(t) = -3.1 + \frac{3.5}{2} t^2 \). The \( x \) component is linear in \( t \) and the \( y \) component is quadratic, which indicates that the path is a parabola. Since the acceleration is constant and only in the \( \hat{\mathbf{j}} \) direction, the motion resembles that of projectile motion, confirming the parabolic path.

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and influenced only by the force of gravity and its initial velocity. Such motion follows a curved trajectory known as a parabola. In the current problem, the particle moves with a combination of linear and quadratic components, making its path a parabola.

• Gravity acts on the trajectory by providing a constant acceleration, which in this example, affects the vertical motion.
• The initial direction and magnitude of velocity determine the shape and distance of the path.

Understanding projectile motion helps predict the path of various bodies, ranging from sports balls to spacecraft, by considering the components of position and velocity as functions of time.

Velocity Function

The velocity function describes how the speed and direction of a particle change over time. In the given exercise, the velocity function is represented by:\(\overrightarrow{\mathbf{v}} = (-2.0 \hat{\mathbf{i}} + 3.5 t \hat{\mathbf{j}}) \, \text{m/s}\).The function consists of two components:

• \(-2.0 \hat{\mathbf{i}}\) - a constant horizontal velocity component.
• \(3.5 t \hat{\mathbf{j}}\) - a vertical velocity component that increases linearly with time.

These components indicate that as time progresses, the particle moves at a steady speed horizontally but accelerates vertically.
The velocity function helps in understanding how the position of the object varies with time, ruling the dynamics of the particle's movement.

Position Function

The position function determines the location of a particle at any given time. In the exercise, we derive it by integrating the velocity function. Initially, the particle starts at the position:\(\overrightarrow{\mathbf{r}} = (1.5 \hat{\mathbf{i}} - 3.1 \hat{\mathbf{j}}) \, \text{m}\).As the velocity function integrates, each component alters the initial position based on time, resulting in the position function:\[\overrightarrow{\mathbf{r}}(t) = (1.5 - 2.0 t)\hat{\mathbf{i}} + \left(-3.1 + \frac{3.5}{2} t^2\right)\hat{\mathbf{j}}.\]

• The horizontal position decreases linearly with time.
• The vertical position changes quadratically due to integration, defining the parabolic path.

This function allows us to predict the exact location of the particle at any point in time, crucial for understanding motion.

Acceleration

Acceleration is a fundamental concept representing how velocity changes with time. It is the derivative of the velocity function. In the problem, the acceleration function is:\(\overrightarrow{\mathbf{a}} = (0 \hat{\mathbf{i}} + 3.5 \hat{\mathbf{j}}) \, \text{m/s}^2\).This indicates:

• There is no horizontal acceleration, meaning no change in horizontal velocity.
• Vertical acceleration is constant, aligning with typical projectile motion due to gravity.

Understanding acceleration gives insight into how forces affect motion, making it easier to predict future dynamics of any moving particle in similar contexts.