Problem 72 - Calculus 3

Question

An object starts from rest at point \(P(1,2,0)\) and moves with an acceleration of \(\mathbf{a}(t)=\mathbf{j}+2 \mathbf{k},\) where \(\|\mathbf{a}(t)\|\) is measured in feet per second per second. Find the location of the object after \(t=2 \mathrm{sec} .\) \(\|\mathbf{a}(t)\|\) is measured in feet per second per second. Find the location of the object after \(t=2 \mathrm{sec} .\)

Step-by-Step Solution

Answer

The location of the object after 2 seconds is \( (1, 4, 4) \).

Step 1: Identify Initial Conditions

The object starts from rest, which means its initial velocity \( \mathbf{v}(0) \) is \( \mathbf{0} \). The initial position of the object is given as \( P(1,2,0) \), so initially at time \( t=0 \), \( \mathbf{r}(0) = \langle 1, 2, 0 \rangle \).

Step 2: Integrate Acceleration to Find Velocity

The acceleration vector is given by \( \mathbf{a}(t) = \mathbf{j} + 2\mathbf{k} \). To find the velocity, we integrate the acceleration with respect to time:\[\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int (0 \mathbf{i} + 1 \mathbf{j} + 2 \mathbf{k}) \, dt = C_1 \mathbf{i} + t \mathbf{j} + 2t \mathbf{k}.\]Given initial velocity \( \mathbf{v}(0) = \mathbf{0} \), we solve for \( C_1 \) to get \( C_1 = 0 \). Thus, the velocity is:\[\mathbf{v}(t) = t \mathbf{j} + 2t \mathbf{k}.\]

Step 3: Integrate Velocity to Find Position

Now integrate the velocity function to find the position:\[\mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int (0 \mathbf{i} + t \mathbf{j} + 2t \mathbf{k}) \, dt = C_2 \mathbf{i} + \frac{t^2}{2} \mathbf{j} + t^2 \mathbf{k}.\]Using the initial position \( \mathbf{r}(0) = \langle 1, 2, 0 \rangle \), we find \( C_2 = 1 \) for \( \mathbf{i} \), and adjust the constants for the \( \mathbf{j} \) and \( \mathbf{k} \) components to match initial conditions. Substituting gives:\[\mathbf{r}(t) = \langle 1, \frac{t^2}{2} + 2, t^2 \rangle.\]

Step 4: Calculate Position at t=2 Seconds

Substitute \( t = 2 \) into the position equation to find:\[\mathbf{r}(2) = \langle 1, \frac{2^2}{2} + 2, (2)^2 \rangle = \langle 1, 2 + 2, 4 \rangle = \langle 1, 4, 4 \rangle.\]Thus, the position of the object at \( t = 2 \) seconds is \( \langle 1, 4, 4 \rangle \).

Key Concepts

Acceleration

Acceleration is a fundamental concept in understanding motion. It describes the rate of change of velocity of an object. In this exercise, the acceleration vector is given as \( \mathbf{a}(t) = \mathbf{j} + 2 \mathbf{k} \), which indicates that there is no acceleration in the \( \mathbf{i} \) (or x) direction. Instead, the object accelerates at a constant rate of 1 unit in the \( \mathbf{j} \) (or y) direction and 2 units in the \( \mathbf{k} \) (or z) direction.

To find velocity or position from acceleration, it's essential to integrate the acceleration vector. Knowing how acceleration affects movement helps us understand more complex motion such as curves or other pathways in space. Simply put, acceleration tells us how quickly the object speeds up or slows down in various directions.

Velocity

Velocity is the rate at which an object's position changes with time. It is a vector quantity, which means it has both magnitude and direction. From the problem, after integrating the acceleration, we find the velocity vector to be \( \mathbf{v}(t) = t \mathbf{j} + 2t \mathbf{k} \). This equation shows how velocity changes over time for each direction.

Notice that the velocity does not contain an \( \mathbf{i} \) component, reflecting no motion in the x-direction. The y-component, \( t \mathbf{j} \), means that the velocity in the y-direction increases linearly with time. Similarly, the velocity in the z-direction, \(2t \mathbf{k}\), also changes linearly over time but is twice as fast as in the y-direction. Understanding the velocity vector helps in predicting future positions of the object.

Position

Position describes the specific point where an object is located in space at a given time. In vector form, it is represented by \( \mathbf{r}(t) \). In this exercise, the position function is derived from the velocity function through integration, resulting in \( \mathbf{r}(t) = \langle 1, \frac{t^2}{2} + 2, t^2 \rangle \).

Each component of this position vector represents a coordinate direction:

• The x-component remains constant at 1, indicating no movement along the x-axis.
• The y-component, \( \frac{t^2}{2} + 2 \), implies that as time progresses, the position increases quadratically with respect to time starting from 2.
• The z-component \( t^2 \) reflects its quadratic increase over time, beginning from a rest position at 0.

Knowing how to express and calculate position is key to describing where an object is at any moment.

Integration

Integration is the mathematical process used to find quantities such as velocity or position from known rates of change, like acceleration. In this exercise, integration was crucial in transitioning from acceleration to velocity, and then from velocity to position.

By integrating the acceleration vector \( \mathbf{a}(t) = \mathbf{j} + 2 \mathbf{k} \), we obtained the velocity vector \( \mathbf{v}(t) = t \mathbf{j} + 2t \mathbf{k} \). A second integration of the velocity vector gives the position vector \( \mathbf{r}(t) = \langle 1, \frac{t^2}{2} + 2, t^2 \rangle \).

Integration helps in reconstructing a global picture of motion from local changes, such as recovering an object's trajectory from its instantaneous acceleration.

Initial Conditions

Initial conditions specify the starting point of an object, which are critical in solving equations of motion involving integration. Here, the initial conditions given were the initial velocity \( \mathbf{v}(0) = \mathbf{0} \) and initial position \( P(1,2,0) \).

These conditions are used to determine the constant terms that arise during integration:

• The constant from integrating acceleration to velocity was determined to be zero because the object starts from rest.
• The constant from integrating velocity to position was found using the initial position, ensuring the trajectory matches the initial known position.

Initial conditions anchor our calculations in the physical scenario described, ensuring that future calculations remain accurate.