91Ó°ÊÓ

A position vector is a mathematical representation of a point in space relative to an origin. It tells us where a particle is located at a particular time. For our exercise, the position vector is given by the equation \[ \vec{r} = \hat{\mathrm{i}} + 4t^2 \hat{\mathrm{j}} + t \hat{\mathrm{k}} \] This vector has three components:

• Along the x-axis: \( \hat{\mathrm{i}} \)
• Along the y-axis: \( 4t^2 \hat{\mathrm{j}} \)
• Along the z-axis: \( t \hat{\mathrm{k}} \)

Let's break it down:

• The "\( \hat{\mathrm{i}} \)" component remains constant, indicating no movement along the x-axis.
• The "4t^2 \hat{\mathrm{j}}" component shows a quadratic relationship with time, suggesting that the particle accelerates along the y-axis.
• The "\( t \hat{\mathrm{k}} \)" component is linear, denoting a steady movement along the z-axis.

By using this vector, you can find the exact position of the particle in three-dimensional space at any time \( t \).

Velocity reveals how the position of a particle changes over time. Mathematically, it is the first derivative of the position vector with respect to time. This describes not only speed but also direction. For our specific case, the velocity vector \( \vec{v} \) is determined by differentiating each component of the position vector:\[ \vec{v} = \frac{d}{dt}(\hat{\mathrm{i}}) + \frac{d}{dt}(4t^2 \hat{\mathrm{j}}) + \frac{d}{dt}(t \hat{\mathrm{k}}) \]This results in:

• \( \frac{d}{dt}(\hat{\mathrm{i}}) = 0 \hat{\mathrm{i}} \), meaning no change along the x-axis.
• \( \frac{d}{dt}(4t^2 \hat{\mathrm{j}}) = 8t \hat{\mathrm{j}} \), indicating acceleration along the y-axis.
• \( \frac{d}{dt}(t \hat{\mathrm{k}}) = \hat{\mathrm{k}} \), showing constant velocity along the z-axis.

Therefore, the velocity vector as a function of time is:\[ \vec{v} = 8t \hat{\mathrm{j}} + \hat{\mathrm{k}} \]This tells us how fast and in what direction the particle is traveling at any moment.

Acceleration indicates how the velocity of a particle changes with time. It is found by differentiating the velocity vector with respect to time, providing more insights into the particle's motion.For the exercise we're studying, the acceleration vector \( \vec{a} \) is obtained by differentiating the velocity vector components:\[ \vec{a} = \frac{d}{dt}(8t \hat{\mathrm{j}}) + \frac{d}{dt}(\hat{\mathrm{k}}) \]Here's the result of those differentiations:

• \( \frac{d}{dt}(8t \hat{\mathrm{j}}) = 8 \hat{\mathrm{j}} \), showing constant acceleration along the y-axis.
• \( \frac{d}{dt}(\hat{\mathrm{k}}) = 0 \hat{\mathrm{k}} \), indicating no acceleration along the z-axis.

Thus, the acceleration vector can be expressed as:\[ \vec{a} = 8 \hat{\mathrm{j}} \]This means that the particle experiences a consistent acceleration purely in the y-direction, providing a full picture of how its motion evolves over time.