Problem 26 - Physics for Scientists and Engineers with Modern Physics

Question

(II) An object, which is at the origin at time \(t=0,\) has $$ \overline{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} $$ $$ \quad \overline{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2} . \text { Find the position } \overline{\mathbf{r}} $$ where the object comes to rest (momentarily).

Step-by-Step Solution

Answer

The position where the object comes to rest is \( \overline{\mathbf{r}} = -\frac{49}{3} \hat{\mathbf{i}} - \frac{24.5}{3} \hat{\mathbf{j}} \).

Step 1: Understand the Problem

We have an object that starts from the origin and moves with an initial velocity \( \overline{\mathbf{v}}_{0}=(-14.0 \hat{\mathbf{i}}-7.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} \) and constant acceleration \( \overline{\mathbf{a}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}^{2}\). We need to determine the position \( \overline{\mathbf{r}} \) when the object momentarily comes to rest.

Step 2: Determine Time When Object Comes to Rest

An object is momentarily at rest when its velocity is zero. Using the velocity equation \( \overline{\mathbf{v}} = \overline{\mathbf{v}}_{0} + \overline{\mathbf{a}} t \), set \( \overline{\mathbf{v}} = \mathbf{0} \) and solve for \( t \). \[-14.0 + 6.0t = 0 \quad \Rightarrow \quad t = \frac{14.0}{6.0} = \frac{7}{3} \, \text{s}\] \[-7.0 + 3.0t = 0 \quad \Rightarrow \quad t = \frac{7.0}{3.0} = \frac{7}{3} \, \text{s}\] Both components give the same time, \( t = \frac{7}{3} \, \text{s} \), which is when the object is at rest.

Step 3: Calculate Position at Time t

Use the position function for constant acceleration: \( \overline{\mathbf{r}} = \overline{\mathbf{r}}_{0} + \overline{\mathbf{v}}_{0} t + \frac{1}{2} \overline{\mathbf{a}} t^2 \). Since the object starts at the origin, \( \overline{\mathbf{r}}_{0} = \mathbf{0} \).\[ \overline{\mathbf{r}} = ((-14)(\frac{7}{3}) + \frac{1}{2}(6)(\frac{7}{3})^2) \hat{\mathbf{i}} + ((-7)(\frac{7}{3}) + \frac{1}{2}(3)(\frac{7}{3})^2) \hat{\mathbf{j}} \]Calculate each component:For \( \hat{\mathbf{i}} \): \[ = (-14 \times \frac{7}{3} + 3 \times \left(\frac{7}{3}\right)^2) = (-98/3 + 49/3) = -49/3 \]For \( \hat{\mathbf{j}} \): \[ = (-7 \times \frac{7}{3} + 1.5 \times \left(\frac{7}{3}\right)^2) = (-49/3 + 24.5/3) = -24.5/3 \]

Step 4: Finalize the Position

Combine the results from each component to get the position vector:\[ \overline{\mathbf{r}} = -\frac{49}{3} \hat{\mathbf{i}} - \frac{24.5}{3} \hat{\mathbf{j}} \]

Key Concepts

Initial Velocity

Initial velocity refers to the speed and direction an object has at the very start of its journey. In physics, this is typically denoted by the symbol \( \overline{\mathbf{v}}_{0} \). It sets the stage for how the object will move. Understanding initial velocity is crucial since it affects how we calculate the final position and any changes in speed over time.
In our exercise, the initial velocity is given as \(( -14.0 \hat{\mathbf{i}} - 7.0 \hat{\mathbf{j}}) \mathrm{m/s} \). These numbers represent vectors, which means the object has a velocity component moving left (negative direction on the \( \hat{\mathbf{i}} \) axis) and a smaller velocity downward (negative direction on the \( \hat{\mathbf{j}} \) axis).

• To break this down: -14.0 m/s in the x-direction and -7.0 m/s in the y-direction indicates it starts moving diagonally downward.

Learning about initial velocity helps you understand that movement is not always straightforward. It often involves multiple directions and requires considering all possible influences at the start.

Acceleration

Acceleration is the rate at which an object's velocity changes over time. In kinematics, it is described by a vector, just like velocity. Acceleration can result in changes in speed, direction, or both. In the problem, the acceleration is given as \( (6.0 \hat{\mathbf{i}} + 3.0 \hat{\mathbf{j}}) \mathrm{m/s}^2 \).
Acceleration has profound effects on how an object moves. It can cause an object to speed up, slow down, or change direction. These influences depend on the magnitude and direction of the acceleration vector.

• X-direction: The acceleration of 6.0 m/s² means the object's velocity is increasing each second to the right on the x-axis.
• Y-direction: Similarly, the 3.0 m/s² acceleration affects the object upward on the y-axis.

By understanding acceleration, we learn how forces change the motion of objects over time, leading to a wide range of possible trajectories.

Position Vector

The position vector describes the location of an object in space relative to a reference point, typically the origin. This vector changes as the object moves over time and is key to determining where an object ends up. In physics, the position vector is often denoted by \( \overline{\mathbf{r}} \).
In the exercise, we derive the position vector at the time the object comes to rest using the initial conditions and the equations of motion. Starting at the origin \( \overline{\mathbf{r}}_{0} = (0,0) \), the final position vector is calculated after determining the time for zero velocity. The math involved may look like:

• The calculation in the \( \hat{\mathbf{i}} \) direction: incorporating time and acceleration to get \( -49/3 \hat{\mathbf{i}} \)
• The calculation in the \( \hat{\mathbf{j}} \) direction: similarly, results in \( -24.5/3 \hat{\mathbf{j}} \)
This results in a position vector \( \overline{\mathbf{r}} = -49/3 \hat{\mathbf{i}} - 24.5/3 \hat{\mathbf{j}} \), indicating final position components. This tells us exactly where the object is in two-dimensional space when it halts momentarily.

Resting Point

The resting point of an object in kinematics refers to the position where the object temporarily stops all movement. This concept is vital when examining how objects move and how we calculate their stopping positions.
Finding the resting point involves calculating when an object's velocity is zero and using its motion equations to determine its coordinates at that moment.

• In the exercise: The velocity formula \( \overline{\mathbf{v}} = \overline{\mathbf{v}}_{0} + \overline{\mathbf{a}} t \) is used, setting \( \overline{\mathbf{v}} = \mathbf{0} \) to solve for time \( t \).
• Once the time to stop is found, the position vector equation \( \overline{\mathbf{r}} = \overline{\mathbf{r}}_{0} + \overline{\mathbf{v}}_{0} t + \frac{1}{2} \overline{\mathbf{a}} t^2 \) finds where it rests.

In our case, the object rest momentarily at \( t = \frac{7}{3} \) seconds, leading to a calculated resting point \( \overline{\mathbf{r}} = -49/3 \hat{\mathbf{i}} - 24.5/3 \hat{\mathbf{j}} \).
Understanding resting points is important for analyzing the motion and designing systems that account for stops, like vehicle brakes or machinery.