Question

When appropriate, consider using the special formulas derived in Exercises 50 and 55 in the remaining exercises. Same paths, different velocity The position functions of objects \(A\) and \(B\) describe different motion along the same path, for \(t \geq 0.\) a. Sketch the path followed by both \(A\) and \(B\). b. Find the velocity and acceleration of \(A\) and \(B\) and discuss the differences. c. Express the acceleration of \(A\) and \(B\) in terms of the tangential and normal components and discuss the differences. $$A: \mathbf{r}(t)=\langle 1+2 t, 2-3 t, 4 t\rangle, B: \mathbf{r}(t)=\langle 1+6 t, 2-9 t, 12 t\rangle$$

Short answer

Question: Sketch the path followed by objects A and B, and discuss the differences in their velocity and acceleration components. Answer: Objects A and B both travel along a linear path, with object B moving faster due to its higher velocity components. While both objects have constant velocities, their acceleration components (tangential and normal) are zero, indicating that their speeds and directions do not change during their motion along the path.

Step-by-step solution

Finding the parametric equations for both objects A and B

For both objects, we are given position functions in vector form. The position functions for A and B can be rewritten in terms of their parametric equations: For Object A: \(x(t) = 1 + 2t\) \(y(t) = 2 - 3t\) \(z(t) = 4t\) For Object B: \(x(t) = 1 + 6t\) \(y(t) = 2 - 9t\) \(z(t) = 12t\)

Analyzing the path of objects A and B

The parametric equations indicate that both objects are traveling along linear paths and have constant velocity components. Since the coefficients of the \(x, y, z\) components of both objects A and B are directly proportional, the objects will be traveling along the same path. However, object B will be traveling faster due to the higher coefficients for each component.

Finding the velocity and acceleration of objects A and B

Velocity is the first derivative of the position function, and acceleration is the second derivative of the position function. For object A: Velocity: \(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle 2, -3, 4 \rangle\) Acceleration: \(\mathbf{a}(t) = \frac{d^2\mathbf{r}(t)}{dt^2} = \langle 0, 0, 0 \rangle\) For object B: Velocity: \(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle 6, -9, 12 \rangle\) Acceleration: \(\mathbf{a}(t) = \frac{d^2\mathbf{r}(t)}{dt^2} = \langle 0, 0, 0 \rangle\) Both objects A and B have constant velocities and zero acceleration. The difference between the velocities of A and B indicate that object B is traveling faster than object A.

Finding the tangential and normal components of acceleration

Since the acceleration of both objects A and B is zero, their tangential and normal components of acceleration are also zero. The tangential acceleration \(a_T\) is defined as the derivative of the speed with respect to time, while the normal acceleration \(a_N\) is given by the formula \(a_N = \frac{||\mathbf{v}(t) \times \mathbf{a}(t) ||}{||\mathbf{v}(t)||^2}\). For both objects A and B: Tangential acceleration \(a_T\): \(0\) Normal acceleration \(a_N\): \(0\)

Discussing the differences in acceleration components

In this exercise, the acceleration components (tangential and normal) of both objects A and B are zero. This means that neither object is experiencing any change in speed or direction during their motion along the path. This is consistent with their velocity vectors, which indicate constant velocities in a straight path. In summary, both objects A and B travel along the same path, but object B moves faster due to a higher velocity. Both objects also have zero acceleration, indicating that their velocities remain constant throughout their motion.

Key concepts

Velocity and Acceleration

Velocity and acceleration are fundamental concepts in physics when analyzing motion. Velocity describes the rate at which an object changes its position, and it is a vector quantity, meaning it has both magnitude and direction. Mathematically, the velocity is given by the first derivative of the position function with respect to time, \(\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}\).

Acceleration, on the other hand, is the rate at which the velocity changes over time. It is also a vector and is calculated as the derivative of the velocity function, or the second derivative of the position function:\[\mathbf{a}(t) = \frac{d^2\mathbf{r}(t)}{dt^2}\].

In the given exercise:

• For object A, the velocity is \(\mathbf{v}(t) = \langle 2, -3, 4 \rangle\).
• For object B, the velocity is \(\mathbf{v}(t) = \langle 6, -9, 12 \rangle\).
• The acceleration for both objects A and B is \(\mathbf{a}(t) = \langle 0, 0, 0 \rangle\), indicating constant velocity.

The absence of acceleration further confirms that both objects maintain their speed and direction along their respective paths.

Tangential and Normal Components

When examining motion, especially in vector form, it's useful to decompose acceleration into tangential and normal components. The tangential acceleration is related to how the speed of the object changes, while the normal component is connected to how the direction changes.

• Tangential Acceleration (\(a_T\)): Defined as the derivative of the speed with respect to time. It indicates changes in the magnitude of velocity.
• Normal Acceleration (\(a_N\)): Calculated as \[a_N = \frac{||\mathbf{v}(t) \times \mathbf{a}(t) ||}{||\mathbf{v}(t)||^2}\], representing how much the direction of the velocity vector changes with respect to time.

For both objects A and B:- Tangential acceleration is zero since their speeds are constant.- Normal acceleration is zero as there is no curvature in their straight-line paths.
This analysis reveals that neither object changes speed or direction along their trajectory.

Constant Velocity

Constant velocity signifies that an object moves with a steady speed and direction over time. In vector terms, a constant velocity implies that the magnitude and direction of velocity do not change.

In the exercise, both objects A and B demonstrate constant velocity. This is derived from their position functions where the velocities:

• For object A: \(\langle 2, -3, 4 \rangle\)
• For object B: \(\langle 6, -9, 12 \rangle\)

remain unchanged throughout their motion.

With zero acceleration for both objects, their velocities stay the same along their straight line path. Object B moves faster along the path compared to object A, which we know by comparing their velocity vectors' magnitudes. The movement of these objects without speed changes creates smooth and predictable paths.

Vector Calculus

Vector calculus is a branch of mathematics particularly useful in understanding motion in multi-dimensional spaces. It deals with vector fields and functions to model physical phenomena such as electromagnetism and fluid dynamics.

In this exercise, vector calculus is employed to:

• Describe the motion of objects using vector-valued functions (position functions \(\mathbf{r}(t)\)).
• Calculate derivatives to find velocity and acceleration vectors.
• Analyze motion by breaking it down into simpler components (such as tangential and normal components).

Understanding vector calculus allows us to precisely determine how objects move through space and time. Knowing how to derive and interpret vectors gives us a clearer picture of both the objects' paths and their relative speeds and directions.