Question

The position vector \(r\) describes the path of an object moving in space. Find the velocity, speed, and acceleration of the object. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j}+2 t^{3 / 2} \mathbf{k} $$

Short answer

The velocity of the object is \(\mathbf{v}(t) = 2t \mathbf{i} + \mathbf{j} + 3t^{1/2} \mathbf{k}\), the speed is \(\sqrt{(2t)^2 + 1^2 + (3t^{1/2})^2}\), and the acceleration is \(\mathbf{a}(t) = 2 \mathbf{i} + 0 \mathbf{j} + \frac{3}{2}t^{-1/2} \mathbf{k}\).

Step-by-step solution

Compute the Velocity

The velocity \(\mathbf{v}(t)\) of the object is the derivative of the position vector \(\mathbf{r}(t)\). Derive each component of \(\mathbf{r}(t)\) with respect to time \(t\): \[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 2t \mathbf{i} + \mathbf{j} + 3t^{1/2} \mathbf{k} \]

Compute the Speed

The speed of the object is the magnitude of the velocity vector \(\mathbf{v}(t)\). It is computed as the square root of the sum of the squares of the components: \[ \text{speed} = |\mathbf{v}(t)| = \sqrt{(2t)^2 + 1^2 + (3t^{1/2})^2} \]

Compute the Acceleration

The acceleration \(\mathbf{a}(t)\) of the object is the derivative of the velocity vector \(\mathbf{v}(t)\). Derive each component of \(\mathbf{v}(t)\) with respect to time \(t\): \[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 2 \mathbf{i} + 0 \mathbf{j} + \frac{3}{2}t^{-1/2} \mathbf{k} \]

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with the differentiation and integration of vector fields, primarily in 3-dimensional Euclidean space. A vector field is a map that assigns each point a vector. Think of it like a weather map showing wind speed and direction at various locations. When we apply the principles of vector calculus to physics, we can describe the motion of objects with vectors that represent quantities such as position, velocity, and acceleration.

In our example, the position vector \( \mathbf{r}(t) = t^{2} \mathbf{i} + t \mathbf{j} + 2t^{3/2} \mathbf{k} \) gives us the location of an object at any time \( t \). This allows us to visualize the object's path through space over time, and by applying vector calculus, we can further analyze its motion to find velocity and acceleration which are critical to understanding how the object moves.

Velocity Vector

A velocity vector provides a way to express both the speed and direction of a moving object. It's the rate at which an object’s position changes with time. To obtain the velocity vector for an object moving in space, as we have in our problem, we take the time derivative of the position vector. Deriving \( \mathbf{r}(t) \) with respect to time yields the velocity vector \( \mathbf{v}(t) \).

For our example, the velocity vector is \( \mathbf{v}(t) = 2t \mathbf{i} + \mathbf{j} + 3t^{1/2} \mathbf{k} \). This equation gives us a full picture of how the object's speed and direction change over time.

Position Vector

The position vector indicates an object's location in space relative to a reference point, often the origin of the coordinate system. Usually denoted as \( \mathbf{r}(t) \), it's a function of time when describing a path of motion. The position vector consists of components in the \( x \), \( y \), and \( z \) directions, represented by the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) respectively.

In our exercise, \( \mathbf{r}(t) = t^{2} \mathbf{i} + t \mathbf{j} + 2t^{3/2} \mathbf{k} \) tells us where the object is at any given time. By examining this vector, we gain insight into the object's trajectory.

Speed Magnitude

The speed magnitude, often referred to simply as 'speed', is the scalar value indicating how fast an object is moving. It's the magnitude of the velocity vector — calculated by taking the square root of the sum of the squares of the velocity vector’s components, which is derived from the Pythagorean theorem applied in three dimensions.

Following the given solution, we calculate the speed as \( \text{speed} = \sqrt{(2t)^2 + 1^2 + (3t^{1/2})^2} \). Unlike the velocity vector, speed doesn't provide direction information; it's purely a measure of how rapidly the object is covering distance over time.

Time Derivative

A time derivative is a measure of how a function changes as time progresses. It's the rate of change of a physical quantity and is foundational in calculus, particularly when analyzing motion. For functions that denote physical quantities tied to an object's movement, such as position, the time derivative gives us the velocity.

We derive \( \mathbf{r}(t) \) with respect to time to obtain the velocity, and similarly, to find the acceleration, we compute the time derivative of the velocity vector \( \mathbf{v}(t) \). This process is what enables us to see changes in velocity and therefore deduce acceleration.

Acceleration Vector

The acceleration vector describes how the velocity of an object changes with time. It conveys both the rate at which the object is speeding up or slowing down and the change in direction of the movement. To find the acceleration vector, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time.

In our given problem, the acceleration vector is found to be \( \mathbf{a}(t) = 2 \mathbf{i} + 0 \mathbf{j} + \frac{3}{2}t^{-1/2} \mathbf{k} \). Just as velocity is the first derivative of position, acceleration is the first derivative of velocity, indicating how dynamic and nuanced motion can be when an object travels through space.