Problem 22 - Thomas Calculus

Question

A particle traveling in a straight line is located at the point \((1,-1,2)\) and has speed 2 at time \(t=0 .\) The particle moves toward the point \((3,0,3)\) with constant acceleration \(2 \mathbf{i}+\mathbf{j}+\mathbf{k}\) Find its position vector \(\mathbf{r}(t)\) at time \(t\) .

Step-by-Step Solution

Answer

The position vector \( \mathbf{r}(t) \) is: \( \langle 1 + \frac{4\sqrt{6}}{6} t + t^2, -1 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2, 2 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2 \rangle \).

Step 1: Understand the Problem

We need to determine the position vector \( \mathbf{r}(t) \) of a particle starting at \( (1,-1,2) \) with a given speed and acceleration at \( t = 0 \). The particle is moving towards another point \( (3,0,3) \). We need to formulate the particle's trajectory over time.

Step 2: Identify Initial Conditions

The particle's initial position vector is \( \mathbf{r}_0 = \langle 1, -1, 2 \rangle \). The initial speed is 2, and since the direction of travel initially is towards \( (3,0,3) \), we calculate the initial velocity vector \( \mathbf{v}_0 \).

Step 3: Calculate Direction Vector

To move towards point \( (3,0,3) \), the direction vector is \( \langle 3-1, 0+1, 3-2 \rangle = \langle 2, 1, 1 \rangle \). Normalize this direction to get \( \widehat{\mathbf{d}} = \frac{1}{\sqrt{6}} \langle 2, 1, 1 \rangle \).

Step 4: Determine Initial Velocity Vector

With the speed 2 in the direction \( \widehat{\mathbf{d}} \), the initial velocity is: \( \mathbf{v}_0 = 2 \times \frac{1}{\sqrt{6}} \langle 2, 1, 1 \rangle = \frac{2}{\sqrt{6}} \langle 2, 1, 1 \rangle = \langle \frac{4}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{2}{\sqrt{6}} \rangle \).

Step 5: Formulate Position Function with Constant Acceleration

The acceleration vector is given as \( \mathbf{a} = \langle 2, 1, 1 \rangle \). The position function for time \( t \) is \( \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \).

Step 6: Substitute and Simplify for \( \mathbf{r}(t) \)

Substitute \( \mathbf{r}_0 = \langle 1, -1, 2 \rangle \), \( \mathbf{v}_0 = \langle \frac{4\sqrt{6}}{6}, \frac{2\sqrt{6}}{6}, \frac{2\sqrt{6}}{6} \rangle \), and \( \mathbf{a} = \langle 2, 1, 1 \rangle \) into \( \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \). Thus, \[ \mathbf{r}(t) = \langle 1, -1, 2 \rangle + \langle \frac{4\sqrt{6}}{6} t, \frac{2\sqrt{6}}{6} t, \frac{2\sqrt{6}}{6} t \rangle + \langle t^2, \frac{1}{2}t^2, \frac{1}{2}t^2 \rangle \].

Step 7: Final Position Vector

Final position vector is: \[ \mathbf{r}(t) = \langle 1 + \frac{4\sqrt{6}}{6} t + t^2, -1 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2, 2 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2 \rangle \].

Key Concepts

Constant Acceleration

When dealing with the motion of particles or objects, constant acceleration is a key concept. It means that the acceleration, or the rate of change of velocity, remains the same throughout the observed period. This assumption simplifies many calculations, as the acceleration does not vary with time, unlike real-world scenarios where forces acting on an object might change.
In this particular problem, the particle experiences a constant acceleration defined by the vector \( \mathbf{a} = \langle 2, 1, 1 \rangle \). This vector indicates the rate at which the particle's velocity changes in each dimension (x, y, and z).

• The x-component accelerates by 2 units per second squared.
• The y-component accelerates by 1 unit per second squared.
• The z-component accelerates by 1 unit per second squared.

This uniform acceleration adds neatly into our equations, allowing us to predict the particle's future state with ease.

Initial Velocity

Understanding initial velocity is critical when analyzing motion. It is the velocity at which a particle or object begins its journey. When a particle's direction and speed are uncertain, we can use initial position and direction information to infer this preliminary velocity.
For our exercise, the initial velocity \( \mathbf{v}_0 \) is calculated by examining the direction the particle is supposed to take. The vector towards the new position \((3,0,3)\) from the current position \((1,-1,2)\) is derived and then used to compute the initial velocity.

Calculating Initial Velocity

1. **Direction Vector:** \( \langle 2, 1, 1 \rangle \) is obtained from the difference between destination and start point.
2. **Normalize:** By dividing this vector by its length, we get the unit direction vector \( \widehat{\mathbf{d}} = \frac{1}{\sqrt{6}} \langle 2, 1, 1 \rangle \).
3. **Speed Apply:** Using the initial speed of 2 units, we multiply by \( \widehat{\mathbf{d}} \) to get \( \mathbf{v}_0 = \langle \frac{4}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{2}{\sqrt{6}} \rangle \).
This gives the complete initial velocity, both in magnitude and direction.

Particle Motion

Particle motion in physics focuses on how a small object moves in space. Understanding and predicting motion require grasping not only the initial conditions but also how forces like acceleration affect it over time.
In a straight-line motion, like in this task, mathematical models describe how position changes due to known forces. Initial conditions, such as position and velocity, combined with external factors like constant acceleration, provide a comprehensive picture of the expected movement.

Modeling Particle Motion

The position function plays a key role:

• It foretells the next location using initial settings and known acceleration.
• The formula \( \mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2} \mathbf{a} t^2 \) encapsulates this essence.

Through these components, we see the step-by-step transformation of positional attributes as time progresses. In simpler terms, it shows us where the particle will be at any moment \( t \).
By integrating each part—starting vector, velocity, and acceleration—the full motion timeline becomes apparent.

Trajectory

Trajectory refers to the path that an object follows through space as a function of time. In our specific situation, finding the particle's trajectory involves combining initial position, velocity, and the influence of constant acceleration.

Deriving the Trajectory

To track the precise route of the particle:

• Start with the known starting point: \( \mathbf{r}_0 = \langle 1, -1, 2 \rangle \).
• Add the journey speed impact: \( \mathbf{v}_0 t \).
• Consider the accelerating effect: \( \frac{1}{2} \mathbf{a} t^2 \).

These elements combine harmoniously to mold the final trajectory:\[\mathbf{r}(t) = \langle 1 + \frac{4\sqrt{6}}{6} t + t^2, -1 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2, 2 + \frac{2\sqrt{6}}{6} t + \frac{1}{2} t^2 \rangle\]
The equation lays out the precise positions at any given second, predicting how the particle's path unfolds with time. This mathematical pathway ensures you can follow every twist and turn of its journey with precision.